On putting sin x = t, given integral reduces to
\(\int_{0}^{1} e^{t} d t .[\text { Ans. } e-1]\)

Let \(I=\int \sqrt{2 x-x^{2}} d x\) 
\(=\int \sqrt{-\left(x^{2}-2 x\right)} d x=\int \sqrt{-\left(x^{2}-2 x+1-1\right)} d x\) 
\(=\int \sqrt{-\left((x-1)^{2}-1\right)} d x=\int \sqrt{1^{2}-(x-1)^{2}} d x\) 
\(=\frac{(x-1)}{2} \sqrt{2 x-x^{2}}+\frac{1}{2} \sin ^{-1}(x-1)+C\)

\(\text { Put } \cos x=t \text { . [Ans. } \cos (\cos x)+C]\)

\(I=\int _{ 0 }^{ 1 }{ \frac { 1 }{ \sqrt { 1-x^{ 2 } } } } dx\)
\(=\left[ { sin }^{ -1 }x \right] ^{ 1 }_{ 0 }\)
\(={ sin }^{ -1 }1-{ sin }^{ -1 }0\)
\(=\frac { \pi }{ 2 } -0=\frac { \pi }{ 2 } \)

\(\int \frac{\sin ^{2} x-\cos ^{2} x}{\sin x \cos x} d x=\int(\tan x-\cot x) d x=\log |\sec x|-\log |\sin x|+C\)

0

\(\int \sqrt{\tan x} \cdot \sec ^{2} x d x=\int \sqrt{t} d t=\frac{2}{3} t^{\frac{3}{2}}+C=\frac{2}{3}(\tan x)^{\frac{3}{2}}+C \)

We have, \(\int \frac{x^2+x+1}{(x+1)^2(x+2)} d x\)
Given integrand \(\frac{x^2+x+1}{(x+1)^2(x+2)}\) is a proper rational function.
Now, by using partial fraction,
\(\text { let } \frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2} \)
\(\Rightarrow x^2+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)^2\)
\( \Rightarrow x^2+x+1=A\left(x^2+3 x+2\right)+B(x+2) +C\left(x^2+2 x+1\right)\)
\(\Rightarrow x^2+x+1=(A+C) x^2+(3 A+B+2 C) x +(2 A+2 B+C)\)
On comparing the coefficients of like powers of x from both sides, we get
A+C  =1 
\(\text { and } \quad 3 A+B+2 C =1\)
2 A+ 2 B + C  =1
On solving these equations, we get
A=-2, B=1 and C=3
From Eq. (i), we get
\(\frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{-2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}\)
\(\therefore \int \frac{x^2+x+1}{(x+1)^2(x+2)} d x=-2 \int \frac{1}{x+1} d x +\int \frac{d x}{(x+1)^2}+3 \int \frac{d x}{(x+2)}\)
\(=-2 \log |x+1|-\frac{1}{x+1}+3 \log |x+2|+C\)

Let \( I=\int \frac{e^x}{\sqrt{e^{2 x}-4 e^x-5}} d x\)
\(\text { Put } e^x=t\)
\(\Rightarrow e^x d x=d t\)
\( \therefore \quad I=\int \frac{d t}{\sqrt{t^2-4 t-5}}\)
\(=\int \frac{d t}{\sqrt{(t-2)^2-(3)^2}} \)
\(=\log \left|(t-2)+\sqrt{(t-2)^2-(3)^2}\right|+C\)
\(=\log \left|\left(e^x-2\right)+\sqrt{\left(e^x-2\right)^2-(3)^2}\right|+C\)
\(=\log \left|\left(e^x-2\right)+\sqrt{e^{2 x}-4 e^x-5}\right|+C\)

Let \(I=\int_{\mathrm{II}} x^{2} \cdot \tan ^{-1} x d x\) 
Using integration by parts, taking tan^-1 x as Ist function and x² as IInd function, we get
\(I=\tan ^{-1} x \int x^{2} d x-\int\left(\frac{d}{d x}\left(\tan ^{-1} x\right) \int x^{2} d x\right) d x\) 
\(I=\tan ^{-1} x \cdot \frac{x^{3}}{3}-\int \frac{1}{1+x^{2}} \cdot \frac{x^{3}}{3} d x\) 
\(=\frac{x^{3}}{3} \cdot \tan ^{-1} x-\frac{1}{3} \int \frac{x^{3}}{1+x^{2}} d x\) 
\(=\frac{x^{3}}{3} \tan ^{-1} x-\frac{1}{3} \int\left(x-\frac{x}{x^{2}+1}\right) d x\)  \(\left[\text { dividing } x^{3} \text { by } x^{2}+1\right]\) 
\(=\frac{x^{3}}{3} \tan ^{-1} x-\frac{1}{3} \int x d x+\frac{1}{3} \int \frac{x}{x^{2}+1} d x\) 
\(=\frac{x^{3}}{3} \tan ^{-1} x-\frac{x^{2}}{6}+\frac{1}{6} \int \frac{2 x}{x^{2}+1} d x\) 
\(=\frac{x^{3}}{3} \tan ^{-1} x-\frac{x^{2}}{6}+\frac{1}{6} \log \left|x^{2}+1\right|+C\) 
\(\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+C\right]\)

\(\int { \frac { \sin { x } }{ \sin { 3x } } } dx=\int { \frac { \sin { x } }{ 3\sin { x } -4\sin ^{ 3 }{ x } } } dx\)
\(=\frac { dx }{ 3-4\sin ^{ 2 }{ x } } \)
[Dividing numerator and denominator by cos²x, we get]
\(I=\int { \frac { \sin ^{ 2 }{ x } dx }{ 3\sin { x } -4\tan ^{ 2 }{ x } } } \)
\(=\int { \frac { \sin ^{ 2 }{ x } dx }{ 3(1+\tan ^{ 2 }{ x } )-4\tan ^{ 2 }{ x } } } =\int { \frac { \sec ^{ 2 }{ x } }{ 3-\tan ^{ 2 }{ x } } } .\)
\(Put\ \tan { x } =t\) so that \(\sec ^{ 2 }{ x } dx=dt.\)
\(\therefore I=\int { \frac { dt }{ 3-{ t }^{ 2 } } } =\int { \frac { dt }{ { (\sqrt { 2 } ) }^{ 2 }-{ t }^{ 2 } } } \)
\(=\frac { 1 }{ 2\sqrt { 3 } } \log { \left| \frac { \sqrt { 3 } +t }{ \sqrt { 3 } -t } \right| } +c\)
\(=\frac { 1 }{ 2\sqrt { 3 } } \log { \left| \frac { \sqrt { 3 } +\tan { x } }{ \sqrt { 3 } -\tan { x } } \right| } +c.\)

17

\(5\over16\)

\(5+10\ log{8\over15}+{25\over2}log{6\over5}\)

As \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ 1+cos\left( \frac { \pi }{ 2 } -x \right) } } \)
= \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 1 }{ 2 } { sec }^{ 2 }\left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) dx } \)
\(=\frac { 1 }{ 2 } .{ \left[ \frac { tan\left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) }{ -\frac { 1 }{ 2 } } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(=-tan\left( \frac { \pi }{ 4 } -\frac { \pi }{ 4 } \right) +tan\left( \frac { \pi }{ 4 } -0 \right) =1\)

As \(\frac { d }{ dx } log|{ 4 }^{ x }+{ x }^{ 4 }|=\)\(\frac { 1 }{ { 4 }^{ x }+{ x }^{ 4 } } .({ 4 }^{ x }.{ log }_{ e }4+{ 4x }^{ 3 })\)=\(\frac { { 4x }^{ 3 }+{ log }_{ e }4.{ 4 }^{ x } }{ { 4 }^{ x }+{ x }^{ 4 } } \)
⇒λ = log_e4

∫sec²(7 – 4x)dx =\(\frac { tan(7-4x) }{ -4 } +C=-\frac { 1 }{ 4 } \)tan(7 - 4x) + C

As \(\frac { d }{ dx } \left( \frac { { 2 }^{ x } }{ { log }_{ e }2 } \right) \)
\(=\frac { 1 }{ { log }_{ e }2 } .{ 2 }^{ x }.{ log }_{ e }2 ={ 2 }^{ x }\)

Let \(I =\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} d x=\int \frac{\sin x-\cos x}{\sqrt{1+\sin 2 x-1}} d x\)
\(=\int \frac{\sin x-\cos x}{\sqrt{\sin ^2 x+\cos ^2 x+2 \sin x \cos x-1}} d x \)
\(=\int \frac{\sin x-\cos x}{\sqrt{(\sin x+\cos x)^2-1}} d x\)
Put \(\sin x+\cos x=t \Rightarrow(\cos x-\sin x) d x=d t\)
\(\therefore  I=\int  \frac{-d t}{\sqrt{t^2-1}}=-\log \left|t+\sqrt{t^2-1}\right|+C\)
\( {\left[\because \int \frac{d x}{\sqrt{x^2-a^2}}=\log \left|x+\sqrt{x^2-a^2}\right|+C\right] }\)
\(\Rightarrow I=  -\log \mid(\sin x+\cos x) +\sqrt{(\sin x+\cos x)^2-1} \mid+C\)
\([\because t=\sin x+\cos x] \)
\(=  -\log \mid(\sin x+\cos x) +\sqrt{\sin ^2 x+\cos ^2 x+2 \sin x \cos x-1} \mid+C\)
\(=  -\log |(\sin x+\cos x)+\sqrt{\sin 2 x}|+C\)

Let \( I=\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x\)
\(=\int \frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \alpha-1\right)}{\cos x-\cos \alpha} d x \)
\(=\int \frac{2\left(\cos ^2 x-\cos ^2 \alpha\right)}{(\cos x-\cos \alpha)} d x \)
\(=\int \frac{2(\cos x-\cos \alpha)(\cos x+\cos \alpha)}{(\cos x-\cos \alpha)} d x\) \( {\left[\because \cos ^2 \theta-1\right] } \)
\(=\int 2(\cos x+\cos \alpha) d x \)
\(=2\left[\int \cos x d x+\cos \alpha \int 1 d x\right]\)
\(\therefore \quad I=2(\sin x+x \cos \alpha)+C\)
 

Let \( I=\int \frac{\sin ^6 x+\cos ^6 x}{\sin ^2 x \cos ^2 x} d x\)
\(\Rightarrow \quad I=\int \frac{\left(\sin ^2 x\right)^3+\left(\cos ^2 x\right)^3}{\sin ^2 x \cos ^2 x} d x\)
\(=\int \frac{\left[\begin{array}{l} \left(\sin ^2 x+\cos ^2 x\right)^3 \\ -3 \sin ^2 x \cos ^2 x\left(\sin ^2 x+\cos ^2 x\right) \end{array}\right]}{\sin ^2 x \cos ^2 x} d x\ \left[\because a^3+b^3=(a+b)^3-3 a b(a+b)\right]\)
\(=\int \frac{(1)^3-3 \sin ^2 x \cos ^2 x}{\sin ^2 x \cos ^2 x} d x\)
\(\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]\)
\(=\int \frac{1}{\sin ^2 x \cos ^2 x} d x-3 \int \frac{\sin ^2 x \cos ^2 x}{\sin ^2 x \cos ^2 x} d x\)
\(=\int \frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x} d x-3 \int 1 d x\)
\(=\int\left[\frac{\sin ^2 x}{\sin ^2 x \cos ^2 x}+\frac{\cos ^2 x}{\sin ^2 x \cos ^2 x}\right] d x-3 \int 1 d x\)
\(=\int\left(\sec ^2 x+\operatorname{cosec}^2 x\right) d x-3 \int 1 d x\)
\(=\int \sec ^2 x d x+\int \operatorname{cosec}^2 x d x-3 \int 1 d x\)
\(=\tan x-\cot x-3 x+C\)

Let  \(I=\int \frac{(2 x-5) e^{2 x}}{(2 x-3)^3} d x=\int \frac{(2 x-3-2) e^{2 x}}{(2 x-3)^3} d x \)
\(=\int \frac{e^{2 x}}{(2 x-3)^2} d x-2 \int \frac{e^{2 x}}{(2 x-3)^3} d x \)
\(=\int e_{11}^{2 x}(2 x-3)^{-2} d x-2 \int e^{2 x}(2 x-3)^{-3} d x\)
\(=\left[\begin{array}{l} (2 x-3)^{-2} \int e^{2 x} d x \\ -\int\left\{\frac{d}{d x}(2 x-3)^{-2} \int e^{2 x} d x\right\} d x \end{array}\right] -2 \int e^{2 x}(2 x-3)^{-3} d x\)
[using integration by parts] 
\( =(2 x-3)^{-2} \frac{e^{2 x}}{2}-\int-2(2 x-3)^{-3} \times 2 \times \frac{e^{2 x}}{2} d x -2 \int e^{2 x}(2 x-3)^{-3} d x\)
\(=\frac{e^{2 x}(2 x-3)^{-2}}{2}+2 \int e^{2 x}(2 x-3)^{-3} d x-2 \int e^{2 x}(2 x-3)^{-3} d x\)
\(=\frac{e^{2 x}(2 x-3)^{-2}}{2}+C\)

Let \(I=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left(5-4 \cos ^2 \theta\right)} d \theta\)
\(=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left[5-4\left(1-\sin ^2 \theta\right)\right]} d \theta \)
\(=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left(5-4+4 \sin ^2 \theta\right)} d \theta\)
\(=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left(1+4 \sin ^2 \theta\right)} d \theta\)
Let \(\sin \theta=t \Rightarrow \cos \theta d \theta=d t\)
Then, \(I=\int \frac{d t}{\left(4+t^2\right)\left(1+4 t^2\right)}\)
Again, let \(\frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)}=\frac{A}{4+t^2}+\frac{B}{1+4 t^2}\)
[by partial fraction]
At \(t=0, \frac{A}{4}+\frac{B}{1}=\frac{1}{4 \times 1} \Rightarrow A+4 B=1\)
At \(t=1, \frac{A}{5}+\frac{B}{5}=\frac{1}{5 \times 5} \Rightarrow 5 A+5 B=1\)
On solving Eqs. (iii) and (iv), we get
\(A=-\frac{1}{15} \text { and } B=\frac{4}{15}\)
On putting \(A=-\frac{1}{15}\) and \(B=\frac{4}{15}\) in Eq. (ii), we get
\(\frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)}=\frac{-\frac{1}{15}}{4+t^2}+\frac{\frac{4}{15}}{1+4 t^2}\)
\(\Rightarrow \quad  \frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)}=\frac{-1}{15\left(4+t^2\right)}+\frac{4}{15\left(1+4 t^2\right)}\)
Now, \( I=\int \frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)} d t \)
\(=\frac{-1}{15} \int \frac{1}{4+t^2} d t+\frac{4}{15} \int \frac{1}{1+4 t^2} d t\)
\(=\frac{-1}{15} \int \frac{1}{2^2+t^2}+\frac{4}{15 \times 4} \int \frac{1}{\left(\frac{1}{2}\right)^2+t^2} d t \)
\(=\frac{-1}{15} \cdot \frac{1}{2} \tan ^{-1} \frac{t}{2}+\frac{1}{15} \cdot \frac{1}{1 / 2} \tan ^{-1} \frac{t}{1 / 2}+C  \quad\left[\because \int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C\right]\)
\(= \frac{-1}{30} \tan ^{-1} \frac{\sin \theta}{2}+\frac{2}{15} \tan ^{-1} 2 \sin \theta+C \quad[\because t=\sin \theta]\)

Let \(I=\int_{0}^{\pi} x \log |\sin x| d x\)   ...(i)  
\(\Rightarrow I=\int_{0}^{\pi}(\pi-x) \log |\sin (\pi-x)| d x\) 
\(=\int_{0}^{\pi}(\pi-x) \log |\sin x| d x\)   ..(ii)
On adding Eqs. (i) and (ii), we get
\(2 I=\pi \int_{0}^{\pi} \log |\sin x| d x\) 
\(\Rightarrow 2 I=2 \pi \int_{0}^{\pi / 2} \log |\sin x| d x\)    ...(iii)
\(\left[\because \int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x, \text { if } f(2 a-x)=f(x)\right]\) 
\(\Rightarrow I=\pi \int_{0}^{\pi / 2} \log |\sin x| d x\)  ...(iv)
\(\Rightarrow I=\pi \int_{0}^{\pi / 2} \log |\sin (\pi / 2-x)| d x\) 
\(\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) 
\(=\pi \int_{0}^{\pi / 2} \log |\cos x| d x\)   ...(v)
On adding Eqs. (iv) and (v), we get
\(2 I=\pi \int_{0}^{\pi / 2}(\log |\sin x|+\log |\cos x|) d x\) 
\(\Rightarrow 2 I I=\pi \int_{0}^{\pi / 2} \log |\sin x \cos x| d x\) 
\(\Rightarrow 2 I=\pi \int_{0}^{\pi / 2} \log \left|\frac{2 \sin x \cos x}{2}\right| d x\) 
[multiply by 2 from numerator and denominator]
\(\Rightarrow 2 I=\pi \int_{0}^{\pi / 2}(\log |\sin 2 x|-\log 2) d x\) 
\(2 I=\pi \int_{0}^{\pi / 2} \log |\sin 2 x| d x-\pi \int_{0}^{\pi / 2} \log 2 d x\) 
\(\Rightarrow 2 I=\pi \int_{0}^{\pi / 2} \log |\sin 2 x| d x-\pi \log 2[x]_{0}^{\pi / 2}\) 
Now, put \(2 x=t \Rightarrow d x=\frac{1}{2} d t\) 
Lower limit When \(x=0, \text { then } t=0\) 
Upper limit When \(x=\frac{\pi}{2}, \text { then } t=\pi\) 
\(\therefore \ 2 I=\frac{\pi}{2} \int_{0}^{\pi} \log |\sin t| d t-\frac{\pi^{2}}{2} \log 2\) 
\(\Rightarrow 2 I=\frac{\pi}{2} \int_{0}^{\pi} \log |\sin x| d x-\frac{\pi^{2}}{2} \log 2\) 
\(\Rightarrow 2 I=I-\frac{\pi^{2}}{2} \log 2\) [from ii]
\(\therefore I=-\frac{\pi^{2}}{2} \log 2=\frac{\pi^{2}}{2} \log \left(\frac{1}{2}\right)\)