Chhaya Mathematics- Indefinite Integral [Part-2]

WB Board Class 12 Mathematics – Integration (SN Dey) Solutions

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1(i) \(\displaystyle \int \dfrac{3x^{2}-4x+5}{\sqrt{x}}\,dx\)

\(\dfrac{3x^{2}-4x+5}{\sqrt{x}} = 3x^{3/2}-4x^{1/2}+5x^{-1/2}\)

\(\Rightarrow \int 3x^{3/2}-4x^{1/2}+5x^{-1/2}\,dx = \frac{6}{5}x^{5/2}-\frac{8}{3}x^{3/2}+10x^{1/2}+C\)

\(\boxed{\frac{6}{5}x^{5/2}-\frac{8}{3}x^{3/2}+10x^{1/2}+C}\)

1(ii) \(\displaystyle \int \dfrac{(x^{2}+1)^{2}}{x^{3}}\,dx\)

\((x^{2}+1)^{2}=x^{4}+2x^{2}+1\Rightarrow\dfrac{x^{4}+2x^{2}+1}{x^{3}}=x+\dfrac{2}{x}+\dfrac{1}{x^{3}}\)

\(\Rightarrow \int (x+\dfrac{2}{x}+\dfrac{1}{x^{3}})\,dx = \frac{x^{2}}{2}+2\ln|x|-\frac{1}{2x^{2}}+C\)

\(\boxed{\frac{x^{2}}{2}+2\ln|x|-\frac{1}{2x^{2}}+C}\)

1(iii) \(\displaystyle \int (a^{2/3}+x^{2/3})^{3}\,dx\)

\((a^{2/3}+x^{2/3})^{3} = a^{2}+3a^{4/3}x^{2/3}+3a^{2/3}x^{4/3}+x^{2}\)

\(\Rightarrow \int(a^{2}+3a^{4/3}x^{2/3}+3a^{2/3}x^{4/3}+x^{2})\,dx\)

\(\ = a^{2}x + 3a^{4/3}\int x^{2/3}\,dx + 3a^{2/3}\int x^{4/3}\,dx + \int x^{2}\,dx\)

\(\Rightarrow a^{2}x + 3a^{4/3}\Big(\frac{3x^{5/3}}{5}\Big) + 3a^{2/3}\Big(\frac{3x^{7/3}}{7}\Big) + \frac{x^{3}}{3} + C\)

\(\Rightarrow \boxed{a^{2}x + \frac{9a^{4/3}x^{5/3}}{5} + \frac{9a^{2/3}x^{7/3}}{7} + \frac{x^{3}}{3} + C}\)

\(\boxed{a^{2}x + \frac{9a^{4/3}x^{5/3}}{5} + \frac{9a^{2/3}x^{7/3}}{7} + \frac{x^{3}}{3} + C}\)

1(iv) \(\displaystyle \int \dfrac{12x^{2}-7x-10}{3x+2}\,dx\)

\((3x+2)(4x-5)=12x^{2}-7x-10\Rightarrow\dfrac{12x^{2}-7x-10}{3x+2}=4x-5\)

\(\Rightarrow \int(4x-5)\,dx=2x^{2}-5x+C\)

\(\boxed{2x^{2}-5x+C}\)

1(v) \(\displaystyle \int \dfrac{x^{3}-6x-9}{x-3}\,dx\)

Divide: \(x^{3}-6x-9=(x-3)(x^{2}+3x+3)\)

\(\Rightarrow \dfrac{x^{3}-6x-9}{x-3}=x^{2}+3x+3\)

\(\Rightarrow \int(x^{2}+3x+3)\,dx=\frac{x^{3}}{3}+\frac{3x^{2}}{2}+3x+C\)

\(\boxed{\frac{x^{3}}{3}+\frac{3x^{2}}{2}+3x+C}\)

1(vi) \(\displaystyle \int \dfrac{x^{6}-1}{x-1}\,dx\)

\(\dfrac{x^{6}-1}{x-1}=x^{5}+x^{4}+x^{3}+x^{2}+x+1\)

\(\Rightarrow \int(x^{5}+x^{4}+x^{3}+x^{2}+x+1)\,dx=\frac{x^{6}}{6}+\frac{x^{5}}{5}+\frac{x^{4}}{4}+\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+C\)

\(\boxed{\frac{x^{6}}{6}+\frac{x^{5}}{5}+\frac{x^{4}}{4}+\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+C}\)

1(vii) \(\displaystyle \int \dfrac{x^{3}-4x^{2}+5x-2}{(x-1)^{2}}\,dx\)

First, we have to factorise the numerator:

\((x^{3}-4x^{2}+5x-2)\)

\(= x^{3}-x^{2}-3x^{2}+3x+2x-2\)

\(= x^{2}(x-1) - 3x(x-1) + 2(x-1)\)

\(= (x-1)(x^{2}-3x+2)\)

\(= (x-1)\{x^{2}-2x - x + 2\}\)

\(= (x-1)\{x(x-2) - 1(x-2)\}\)

\(= (x-1)(x-2)(x-1)\)

\(= (x-1)^{2}(x-2)\)

∴ \(\displaystyle \int \dfrac{x^{3}-4x^{2}+5x-2}{(x-1)^{2}}\,dx\)

\(= \int \dfrac{(x-1)^{2}(x-2)}{(x-1)^{2}}\,dx\)

\(= \int (x-2)\,dx\)

\(= \int x\,dx - 2\int dx\)

\(= \dfrac{x^{2}}{2} - 2x + C\)

\(\boxed{\dfrac{x^{2}}{2} - 2x + C}\)

1(viii) \(\displaystyle \int \dfrac{x^{4}+x^{2}+1}{x^{2}-x+1}\,dx\)

\(x^{4}+x^{2}+1=(x^{2}-x+1)(x^{2}+x+1)\)

\(\Rightarrow \dfrac{x^{4}+x^{2}+1}{x^{2}-x+1}=x^{2}+x+1\)

\(\Rightarrow \int(x^{2}+x+1)\,dx=\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+C\)

\(\boxed{\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+C}\)

1(ix) \(\displaystyle \int \dfrac{2x^{3}-7x+2}{x^{3}+2x^{2}}\,dx\)

Denominator \(=x^{2}(x+2)\)

\(x=-2\Rightarrow 0\Rightarrow (x+2)\) factor of numerator

Divide \(2x^{3}-7x+2=(x+2)(2x^{2}-4x+1)\)

\(\Rightarrow \dfrac{2x^{3}-7x+2}{x^{2}(x+2)}=\dfrac{2x^{2}-4x+1}{x^{2}}=2-\dfrac{4}{x}+\dfrac{1}{x^{2}}\)

\(\Rightarrow \int(2-\dfrac{4}{x}+\dfrac{1}{x^{2}})\,dx=2x-4\ln|x|-\dfrac{1}{x}+C\)

\(\boxed{2x-4\ln|x|-\dfrac{1}{x}+C}\)

2(i) \(\displaystyle \int \dfrac{e^{6\log x}-e^{4\log x}}{e^{3\log x}-e^{\log x}}\,dx\)

\(e^{k\log x}=x^{k}\Rightarrow\dfrac{x^{6}-x^{4}}{x^{3}-x}=\dfrac{x^{4}(x^{2}-1)}{x(x^{2}-1)}=x^{3}\)

\(\Rightarrow \int x^{3}\,dx=\frac{x^{4}}{4}+C\)

\(\boxed{\frac{x^{4}}{4}+C}\)

2(ii) \(\displaystyle \int \frac{e^{2x-1}-e^{1-2x}}{e^{x+2}}\,dx\)

\(\dfrac{e^{2x-1}-e^{1-2x}}{e^{x+2}} = e^{(2x-1)-(x+2)} - e^{(1-2x)-(x+2)}\)

\(\ = e^{x-3} - e^{-3x-1}\)

\(\displaystyle \int\big(e^{x-3}-e^{-3x-1}\big)\,dx = \int e^{x-3}\,dx - \int e^{-3x-1}\,dx\)

\(\displaystyle \int e^{x-3}\,dx = e^{-3}\int e^{x}\,dx = e^{x-3}\)

\(\displaystyle \int e^{-3x-1}\,dx = e^{-1}\int e^{-3x}\,dx = e^{-1}\cdot\frac{e^{-3x}}{-3} = -\tfrac{1}{3}e^{-3x-1}\)

\(\Rightarrow \displaystyle \int \frac{e^{2x-1}-e^{1-2x}}{e^{x+2}}\,dx = e^{x-3} + \tfrac{1}{3}e^{-3x-1} + C\)

\(\boxed{\,e^{x-3} + \tfrac{1}{3}e^{-3x-1} + C\,}\)

3(ii) \(\displaystyle \int \dfrac{27^{1+x}+9^{1-x}}{3^{x}}\,dx\)

\(27^{1+x}=27\cdot3^{3x},\ 9^{1-x}=9\cdot3^{-2x}\)

\(\Rightarrow \dfrac{27\cdot3^{3x}+9\cdot3^{-2x}}{3^{x}}=27\cdot3^{2x}+9\cdot3^{-3x}\)

\(\Rightarrow \int(27\cdot3^{2x}+9\cdot3^{-3x})\,dx =\frac{27\cdot3^{2x}}{2\ln3}-\frac{3\cdot3^{-3x}}{\ln3}+C\)

\(\boxed{\frac{27\cdot3^{2x}}{2\ln3}-\frac{3\cdot3^{-3x}}{\ln3}+C}\)

3(iii) \(\displaystyle \int \dfrac{2^{6x}-1}{2^{2x}-1}\,dx\)

Let \(t=2^{2x}\Rightarrow2^{6x}=t^{3}\)

\(\Rightarrow\dfrac{t^{3}-1}{t-1}=t^{2}+t+1\)

\(dt=2\ln2\,t\,dx\Rightarrow dx=\dfrac{dt}{2\ln2\,t}\)

\(\Rightarrow \int(t^{2}+t+1)\,dx=\frac{1}{2\ln2}\int(t+1+\tfrac{1}{t})\,dt\)

\(\Rightarrow \frac{1}{2\ln2}(\frac{t^{2}}{2}+t+\ln|t|)+C\)

Substitute \(t=2^{2x}\Rightarrow\boxed{\frac{2^{4x}}{4\ln2}+\frac{2^{2x}}{2\ln2}+x+C}\)

\(\boxed{\frac{2^{4x}}{4\ln2}+\frac{2^{2x}}{2\ln2}+x+C}\)

4.Evaluate the following integrals

(4.i) \(\displaystyle \int \cos^{3}x\,dx\)

Use identity \( \cos 3x = 4\cos^{3}x - 3\cos x \)

Hence \( \cos^{3}x = \dfrac{\cos 3x + 3\cos x}{4} \)

\(\displaystyle \int \cos^{3}x\,dx = \frac{1}{4}\int \cos 3x\,dx + \frac{3}{4}\int \cos x\,dx\)

\(\ = \frac{1}{12}\sin 3x + \frac{3}{4}\sin x + C\)

\(\boxed{\;\dfrac{1}{12}\sin 3x + \dfrac{3}{4}\sin x + C\;}\)

(4.ii) \(\displaystyle \int \sin^{4}x\,dx\)

Start: \(\sin^{2}x = \dfrac{1-\cos 2x}{2}\)

Then \(\sin^{4}x = (\sin^{2}x)^{2} = \left(\dfrac{1-\cos 2x}{2}\right)^{2}\)

\(\displaystyle = \dfrac{1 - 2\cos 2x + \cos^{2}2x}{4}\)

Use \(\cos^{2}2x = \dfrac{1+\cos 4x}{2}\)

\(\displaystyle \sin^{4}x = \dfrac{1 - 2\cos 2x + \tfrac{1+\cos4x}{2}}{4}\)

\(\displaystyle = \dfrac{2 - 4\cos2x + 1 + \cos4x}{8} = \dfrac{3 - 4\cos2x + \cos4x}{8}\)

\(\displaystyle \int \sin^{4}x\,dx = \int \dfrac{3 - 4\cos2x + \cos4x}{8}\,dx\)

\(\displaystyle = \frac{3x}{8} - \frac{\sin2x}{4} + \frac{\sin4x}{32} + C\)

\(\boxed{\;\dfrac{3x}{8}-\dfrac{\sin2x}{4}+\dfrac{\sin4x}{32}+C\;}\)

(4.iii) \(\displaystyle \int \cos^{4}x\,dx\)

Start: \(\cos^{2}x = \dfrac{1+\cos 2x}{2}\)

Then \(\cos^{4}x = (\cos^{2}x)^{2} = \left(\dfrac{1+\cos2x}{2}\right)^{2}\)

\(\displaystyle = \dfrac{1 + 2\cos2x + \cos^{2}2x}{4}\)

Use \(\cos^{2}2x = \dfrac{1+\cos4x}{2}\)

\(\displaystyle \cos^{4}x = \dfrac{1 + 2\cos2x + \tfrac{1+\cos4x}{2}}{4}\)

\(\displaystyle = \dfrac{2 + 4\cos2x + 1 + \cos4x}{8} = \dfrac{3 + 4\cos2x + \cos4x}{8}\)

\(\displaystyle \int \cos^{4}x\,dx = \int \dfrac{3 + 4\cos2x + \cos4x}{8}\,dx\)

\(\displaystyle = \frac{3x}{8} + \frac{\sin2x}{4} + \frac{\sin4x}{32} + C\)

\(\boxed{\;\dfrac{3x}{8}+\dfrac{\sin2x}{4}+\dfrac{\sin4x}{32}+C\;}\)

(4.iv) \(\displaystyle \int \dfrac{\cos x+\sin x}{\cos x-\sin x}\,(1-\sin2x)\,dx\)

Identity: \(\sin2x = 2\sin x\cos x\)

Hence \(1-\sin2x = 1 - 2\sin x\cos x\)

Note \((\cos x - \sin x)^{2} = \cos^{2}x - 2\sin x\cos x + \sin^{2}x = 1 - 2\sin x\cos x\)

\(\Rightarrow 1-\sin2x = (\cos x - \sin x)^{2}\)

\(\displaystyle \dfrac{\cos x+\sin x}{\cos x-\sin x}\,(1-\sin2x) = (\cos x+\sin x)(\cos x-\sin x)\)

\((\cos x+\sin x)(\cos x-\sin x) = \cos^{2}x - \sin^{2}x = \cos2x\)

\(\displaystyle \int \cos2x\,dx = \tfrac{1}{2}\sin2x + C\)

\(\boxed{\;\tfrac{1}{2}\sin2x + C\;}\)

(4.v) \(\displaystyle \int \dfrac{2-3\sin x}{\cos^{2}x}\,dx\)

Write \( \dfrac{2-3\sin x}{\cos^{2}x} = 2\sec^{2}x - 3\cdot\dfrac{\sin x}{\cos^{2}x}\)

Note \( \dfrac{\sin x}{\cos^{2}x} = \dfrac{\sin x}{\cos x}\cdot\sec x = \tan x\sec x \)

\(\displaystyle \int 2\sec^{2}x\,dx - 3\int \tan x\sec x\,dx = 2\tan x - 3\sec x + C\)

\(\boxed{\;2\tan x - 3\sec x + C\;}\)

(4.vi) \(\displaystyle \int \dfrac{\sin^{3}x+\cos^{3}x}{\sin x+\cos x}\,dx\)

Factor: \(a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})\)

\(\Rightarrow \dfrac{\sin^{3}x+\cos^{3}x}{\sin x+\cos x}= \sin^{2}x - \sin x\cos x + \cos^{2}x = 1 - \sin x\cos x\)

Use \( \sin x\cos x = \tfrac{1}{2}\sin 2x\)

\(\displaystyle \int (1 - \tfrac{1}{2}\sin 2x)\,dx = x + \tfrac{1}{4}\cos 2x + C\)

\(\boxed{\;x + \dfrac{1}{4}\cos 2x + C\;}\)

(4.vii) \(\displaystyle \int \dfrac{1+\cos x}{1-\cos x}\,dx\)

Identity: \( \dfrac{1+\cos x}{1-\cos x} = \cot^{2}\!\dfrac{x}{2}\)

Also \( \cot^{2}u = \csc^{2}u - 1 \)

\(\displaystyle \int \cot^{2}\!\dfrac{x}{2}\,dx = \int (\csc^{2}\!\dfrac{x}{2} -1)\,dx\)

Since \( \dfrac{d}{dx}\cot\dfrac{x}{2} = -\tfrac{1}{2}\csc^{2}\!\dfrac{x}{2}\), we have

\(\Rightarrow \int \csc^{2}\!\dfrac{x}{2}\,dx = -2\cot\dfrac{x}{2}\)

\(\displaystyle \int \cot^{2}\!\dfrac{x}{2}\,dx = -2\cot\dfrac{x}{2} - x + C\)

\(\boxed{\;-2\cot\frac{x}{2} - x + C\;}\)

(4.viii) \(\displaystyle \int \dfrac{\sin\theta+\cos\theta}{\sqrt{1+\sin2\theta}}\,d\theta\)

Identity: \(1+\sin2\theta=(\sin\theta+\cos\theta)^{2}\)

Assume \(\sin\theta+\cos\theta>0\Rightarrow \sqrt{1+\sin2\theta}=\sin\theta+\cos\theta\)

\(\Rightarrow \dfrac{\sin\theta+\cos\theta}{\sqrt{1+\sin2\theta}} = 1\)

\(\Rightarrow \boxed{\;\theta + C\;}\)

(4.ix) \(\displaystyle \int (\cos^{3}\theta - 3\cos\theta\sin^{2}\theta)\,d\theta\)

Use \( \sin^{2}\theta = 1-\cos^{2}\theta \)

\(\cos^{3}\theta - 3\cos\theta\sin^{2}\theta = \cos^{3}\theta - 3\cos\theta(1-\cos^{2}\theta)\)

\(\ = 4\cos^{3}\theta - 3\cos\theta = \cos 3\theta\)

\(\displaystyle \int \cos 3\theta \,d\theta = \tfrac{1}{3}\sin 3\theta + C\)

\(\boxed{\;\tfrac{1}{3}\sin 3\theta + C\;}\)

(4.x) \(\displaystyle \int \dfrac{dx}{\sin^{2}x\cos^{2}x}\)

Use \( \sin 2x = 2\sin x\cos x \Rightarrow \sin^{2}2x = 4\sin^{2}x\cos^{2}x \)

\(\Rightarrow \dfrac{1}{\sin^{2}x\cos^{2}x} = \dfrac{4}{\sin^{2}2x} = 4\csc^{2}2x\)

\(\displaystyle \int 4\csc^{2}2x\,dx = 4\cdot\Big(-\tfrac{1}{2}\cot 2x\Big) + C = -2\cot 2x + C\)

\(\boxed{\;-2\cot 2x + C\;}\)

(4.xi) \(\displaystyle \int \dfrac{a\sin^{3}x + b\cos^{3}x}{\sin^{2}x\cos^{2}x}\,dx\)

Rewrite numerator over denominator:

\(\dfrac{a\sin^{3}x}{\sin^{2}x\cos^{2}x} = a\cdot\dfrac{\sin x}{\cos^{2}x} = a\,\tan x\sec x\)

\(\dfrac{b\cos^{3}x}{\sin^{2}x\cos^{2}x} = b\cdot\dfrac{\cos x}{\sin^{2}x} = b\,\cot x\csc x\)

\(\displaystyle \int a\tan x\sec x\,dx + \int b\cot x\csc x\,dx = a\sec x - b\csc x + C\)

\(\boxed{\;a\sec x - b\csc x + C\;}\)

(4.xii) \(\displaystyle \int \dfrac{\cos^{2}x - \sin^{2}x}{\sqrt{1+\cos4x}}\,dx\)

\(\cos^{2}x-\sin^{2}x=\cos2x\)

Use \(1+\cos4x = 2\cos^{2}2x\Rightarrow \sqrt{1+\cos4x}=\sqrt{2}\,|\cos2x|\)

Assume \(\cos2x>0\Rightarrow \dfrac{\cos2x}{\sqrt{1+\cos4x}}=\dfrac{1}{\sqrt{2}}\)

\(\Rightarrow \boxed{\;\dfrac{x}{\sqrt{2}} + C\;}\)

(4.xiii) \(\displaystyle \int \tan^{-1}\!\Big(\dfrac{\sin x}{1-\cos x}\Big)\,dx\)

Identity chain:

\(\sin x = 2\sin\dfrac{x}{2}\cos\dfrac{x}{2},\quad 1-\cos x = 2\sin^{2}\dfrac{x}{2}\)

\(\Rightarrow \dfrac{\sin x}{1-\cos x} = \dfrac{2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}{2\sin^{2}\dfrac{x}{2}} = \dfrac{\cos\dfrac{x}{2}}{\sin\dfrac{x}{2}} = \cot\dfrac{x}{2}\)

\(\Rightarrow \tan^{-1}\!\Big(\dfrac{\sin x}{1-\cos x}\Big) = \tan^{-1}\!\big(\cot\tfrac{x}{2}\big)\)

On principal values \(\tan^{-1}(\cot u)=\tfrac{\pi}{2}-u\). With \(u=\tfrac{x}{2}\):

\(\Rightarrow \tan^{-1}\!\big(\cot\tfrac{x}{2}\big)=\tfrac{\pi}{2}-\tfrac{x}{2}\)

\(\displaystyle \int\Big(\tfrac{\pi}{2}-\tfrac{x}{2}\Big)\,dx = \frac{\pi x}{2} - \frac{x^{2}}{4} + C\)

\(\boxed{\;\frac{\pi x}{2} - \frac{x^{2}}{4} + C\;}\)

Show that :

(5.i) Show that \(\displaystyle \int \sqrt{1+\cos x}\,dx = 2\sqrt{2}\,\sin\frac{x}{2} + C\)

Identity: \(1+\cos x = 2\cos^{2}\dfrac{x}{2}\).

Hence \(\displaystyle \sqrt{1+\cos x} = \sqrt{2}\cos\dfrac{x}{2}.\)

So \(\displaystyle \int \sqrt{1+\cos x}\,dx = \sqrt{2}\int \cos\dfrac{x}{2}\,dx.\)

Integrate: \(\displaystyle \int \cos\dfrac{x}{2}\,dx = 2\sin\dfrac{x}{2}.\)

Therefore \(\displaystyle \int \sqrt{1+\cos x}\,dx = 2\sqrt{2}\,\sin\dfrac{x}{2} + C.\)

Hence proved. \(\boxed{\,2\sqrt{2}\,\sin\dfrac{x}{2} + C\,}\)

(5.ii) Show that \(\displaystyle \int \sqrt{1-\cos 2x}\,dx = -\sqrt{2}\,\cos x + C\)

Identity: \(1-\cos 2x = 2\sin^{2}x.\)

Hence \(\displaystyle \sqrt{1-\cos 2x} = \sqrt{2}\sin x.\)

So \(\displaystyle \int \sqrt{1-\cos 2x}\,dx = \sqrt{2}\int \sin x\,dx.\)

Integrate: \(\displaystyle \int \sin x\,dx = -\cos x.\)

Therefore \(\displaystyle \int \sqrt{1-\cos 2x}\,dx = -\sqrt{2}\cos x + C.\)

Hence proved. \(\boxed{\, -\sqrt{2}\cos x + C \,}\)

(5.iii) Show that \(\displaystyle \int \sqrt{1+\sin 2x}\,dx = \sin x - \cos x + C\)

Identity: \(1+\sin 2x = (\sin x + \cos x)^{2}.\)

Hence \(\displaystyle \sqrt{1+\sin 2x} = \sin x + \cos x.\)

So \(\displaystyle \int \sqrt{1+\sin 2x}\,dx = \int (\sin x + \cos x)\,dx.\)

Integrate: \(\displaystyle \int (\sin x + \cos x)\,dx = -\cos x + \sin x.\)

Therefore \(\displaystyle \int \sqrt{1+\sin 2x}\,dx = \sin x - \cos x + C.\)

Hence proved. \(\boxed{\,\sin x - \cos x + C\,}\)

(5.iv) Show that \(\displaystyle \int \sqrt{1+\sin x}\,dx = 2\Big(\sin\frac{x}{2}-\cos\frac{x}{2}\Big) + C\)

Observe: \((\sin\tfrac{x}{2} + \cos\tfrac{x}{2})^{2} = \sin^{2}\tfrac{x}{2} + 2\sin\tfrac{x}{2}\cos\tfrac{x}{2} + \cos^{2}\tfrac{x}{2} = 1 + \sin x.\)

Hence \(\displaystyle \sqrt{1+\sin x} = \sin\tfrac{x}{2} + \cos\tfrac{x}{2}.\)

So \(\displaystyle \int \sqrt{1+\sin x}\,dx = \int\big(\sin\tfrac{x}{2} + \cos\tfrac{x}{2}\big)\,dx.\)

Integrate: \(\displaystyle \int \sin\tfrac{x}{2}\,dx = -2\cos\tfrac{x}{2},\quad \int \cos\tfrac{x}{2}\,dx = 2\sin\tfrac{x}{2}.\)

Therefore \(\displaystyle \int \sqrt{1+\sin x}\,dx = 2\sin\tfrac{x}{2} - 2\cos\tfrac{x}{2} + C.\)

Hence proved. \(\boxed{\,2\Big(\sin\tfrac{x}{2}-\cos\tfrac{x}{2}\Big) + C\,}\)

(v) \( \displaystyle \int \frac{dx}{1+\sin x} = \tan x - \sec x + C \)

Multiply numerator and denominator by \(1-\sin x\):

\( \displaystyle \int \frac{1-\sin x}{\cos^{2}x}\,dx = \int \sec^{2}x\,dx - \int \tan x\sec x\,dx \)

Integrate each term: \( \tan x - \sec x + C \)

Hence proved. \(\boxed{\tan x - \sec x + C}\)

(vi) \( \displaystyle \int \frac{dx}{1-\cos x} = -(\csc x + \cot x) + C \)

Identity: \(1-\cos x = 2\sin^{2}\tfrac{x}{2}\)

⇒ \( \dfrac{dx}{1-\cos x} = \dfrac{1}{2\sin^{2}\tfrac{x}{2}}dx = \tfrac{1}{2}\csc^{2}\tfrac{x}{2}dx \)

Integrate: \( \tfrac{1}{2}\int \csc^{2}\tfrac{x}{2}dx = -\cot\tfrac{x}{2} + C \)

Now \( \cot\tfrac{x}{2} = \csc x + \cot x \)

Hence proved. \(\boxed{-\,(\csc x + \cot x) + C}\)

(vii) \( \displaystyle \int \frac{\cos x\,dx}{\sqrt{1+\sin x}} = 2\Big(\sin\frac{x}{2} + \cos\frac{x}{2}\Big) + C \)

Let \(1+\sin x = 2\sin^{2}\tfrac{x}{2} + 2\sin\tfrac{x}{2}\cos\tfrac{x}{2}\). But \( \sqrt{1+\sin x} = \sin\tfrac{x}{2} + \cos\tfrac{x}{2}. \)

Use \( \cos x = \cos^{2}\tfrac{x}{2} - \sin^{2}\tfrac{x}{2} \)

⇒ Integrand simplifies to \( (\cos\tfrac{x}{2} - \sin\tfrac{x}{2}) \)

\( \displaystyle \int(\cos\tfrac{x}{2}-\sin\tfrac{x}{2})dx = 2\sin\tfrac{x}{2} + 2\cos\tfrac{x}{2} + C \)

Hence proved. \(\boxed{2\Big(\sin\tfrac{x}{2}+\cos\tfrac{x}{2}\Big) + C}\)

(viii) \( \displaystyle \int \sin7x\,\cos3x\,dx = -\Big[\tfrac{1}{20}\cos10x + \tfrac{1}{8}\cos4x\Big] + C \)

Use product identity \( \sin A\cos B = \tfrac{1}{2}[\sin(A+B)+\sin(A-B)] \)

\( \sin7x\cos3x = \tfrac{1}{2}[\sin10x+\sin4x] \)

Integrate: \( \tfrac{1}{2}\int(\sin10x+\sin4x)\,dx = -\tfrac{1}{20}\cos10x - \tfrac{1}{8}\cos4x + C \)

Hence proved. \(\boxed{-\Big(\tfrac{1}{20}\cos10x+\tfrac{1}{8}\cos4x\Big)+C}\)

(ix) \( \displaystyle \int \cos2x\,\cos4x\,dx = \frac{\sin6x}{12} + \frac{\sin2x}{4} + C \)

Use \( \cos A\cos B = \tfrac{1}{2}[\cos(A-B)+\cos(A+B)] \)

\( \cos2x\cos4x = \tfrac{1}{2}[\cos2x+\cos6x] \)

Integrate: \( \tfrac{1}{2}\int(\cos2x+\cos6x)\,dx = \tfrac{1}{4}\sin2x + \tfrac{1}{12}\sin6x + C \)

Hence proved. \(\boxed{\tfrac{\sin6x}{12} + \tfrac{\sin2x}{4} + C}\)

(x) \( \displaystyle \int \cos7x\,\sin4x\,dx = \tfrac{1}{6}\cos3x - \tfrac{1}{22}\cos11x + C \)

Use \( \sin A\cos B = \tfrac{1}{2}[\sin(A+B)+\sin(A-B)] \)

\( \cos7x\sin4x = \tfrac{1}{2}[\sin(11x)-\sin(3x)] \)

Integrate: \( \tfrac{1}{2}\int(\sin11x-\sin3x)\,dx = -\tfrac{1}{22}\cos11x + \tfrac{1}{6}\cos3x + C \)

Hence proved. \(\boxed{\tfrac{1}{6}\cos3x - \tfrac{1}{22}\cos11x + C}\)

(xi) \( \displaystyle \int \sin7x\,\sin3x\,dx = \tfrac{1}{8}\sin4x - \tfrac{1}{20}\sin10x + C \)

Use \( \sin A\sin B = \tfrac{1}{2}[\cos(A-B)-\cos(A+B)] \)

\( \sin7x\sin3x = \tfrac{1}{2}[\cos4x-\cos10x] \)

Integrate: \( \tfrac{1}{2}\int(\cos4x-\cos10x)\,dx = \tfrac{1}{8}\sin4x - \tfrac{1}{20}\sin10x + C \)

Hence proved. \(\boxed{\tfrac{1}{8}\sin4x - \tfrac{1}{20}\sin10x + C}\)

6(i) The slope of a curve at \((x,y)\) is \(2x+1\). If the curve passes through \((-4,2)\), find the equation of the curve.

Given \(\dfrac{dy}{dx} = 2x + 1\).

Integrate: \(y = \int (2x+1)\,dx = x^{2} + x + C\).

Use point \((-4,2)\): \(2 = (-4)^{2} + (-4) + C = 16 - 4 + C = 12 + C\).

So \(C = 2 - 12 = -10\).

Hence the curve: \( \boxed{\,y = x^{2} + x - 10\,} \).

Hence proved.

6(ii) Find the equation of the curve whose slope at any point \((x,y)\) is \(-y\) and which passes through \((2,1)\).

Given \(\dfrac{dy}{dx} = -y\).

Separate variables: \(\dfrac{dy}{y} = -dx\).

Integrate: \(\ln y = -x + C\) ⇒ \(y = Ce^{-x}\).

Use \((2,1)\): \(1 = C e^{-2}\) ⇒ \(C = e^{2}\).

Hence \( \boxed{\,y = e^{2-x}\,} \).

Hence proved.

7 Find a function whose differential is \(\sin^{2}x\cos^{2}x\,dx\).

We integrate \( \sin^{2}x\cos^{2}x\,dx \).

Use \( \sin^{2}x\cos^{2}x = \dfrac{1}{4}\sin^{2}2x \) and \( \sin^{2}2x = \dfrac{1-\cos4x}{2} \).

So \( \sin^{2}x\cos^{2}x = \dfrac{1}{8}\big(1-\cos4x\big).\)

Integrate: \( \int \sin^{2}x\cos^{2}x\,dx = \tfrac{1}{8}\int(1-\cos4x)\,dx = \tfrac{x}{8} - \tfrac{\sin4x}{32} + C.\)

A function with given differential is \( \boxed{\,F(x)=\dfrac{x}{8} - \dfrac{\sin4x}{32} + C\,} \).

Hence proved.

8 If \( \dfrac{dy}{dx} = 3\sqrt{x} - \dfrac{1}{\sqrt{x}} \), find \(y\) as a function of \(x\); given \(y=12\) when \(x=4\).

Rewrite: \( \dfrac{dy}{dx} = 3x^{1/2} - x^{-1/2}.\)

Integrate: \( y = \int(3x^{1/2} - x^{-1/2})\,dx = 3\cdot \frac{2}{3}x^{3/2} - 2x^{1/2} + C = 2x^{3/2} - 2x^{1/2} + C.\)

Use \(y(4)=12\): \(12 = 2(4)^{3/2} - 2(4)^{1/2} + C = 2\cdot 8 - 2\cdot 2 + C = 16 - 4 + C = 12 + C\).

So \(C = 0\).

Hence \( \boxed{\,y = 2x^{3/2} - 2x^{1/2}\,} \).

Hence proved.

Next Topics in Indefinite Integration

Indefinite Integral – Very Short Answer Type [2 Marks] Indefinite Integral – Short Answer Type [4 Marks] Indefinite Integral – Long Answer Type [5 Marks]

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