Ex 7.2 integration ncert solution maths class 12

 EXERCISE 7.2(Integration)

Integrate the functions in Exercises 1 to 37:(Ex 7.2 integration ncert solution maths class 12)

Question 1: \frac{2 x}{1+x^2}

Solution: Let I = \int\frac{2x}{1+x^2}dx

Let 1+x^2=t

\therefore 2 x d x=d t

\Rightarrow \int \frac{2 x}{1+x^2} d x=\int \frac{1}{t} d t

=\log |t|+C

=\log \left|1+x^2\right|+C

=\log \left(1+x^2\right)+C

where \mathrm{C} is an arbitrary constant.

Question 2:  \frac{(\log x)^2}{x}

Solution: Let I=\frac{(\log x)^2}{x}

Let \log x=t

\therefore \frac{1}{x} d x=d t

\Rightarrow \int \frac{(\log |x|)^2}{x} d x=\int t^2 d t

=\frac{t^3}{3}+C

=\frac{(\log |x|)^3}{3}+C

Question 3: Integrate \frac{1}{x+x \log x}

Solution: Let I =\int  \frac{1}{x+x \log x}dx

I=\int \frac{1}{x(1+\log x)}

\text { Let } 1+\log x=t

\therefore \frac{1}{x} d x=d t

\Rightarrow \int \frac{1}{x(1+\log x)} d x=\int\frac{1}{t} d t

=\log |t|+C

=\log |1+\log x|+C

Question 4: Integrate \sin x \cdot \sin (\cos x)

Solution: Let I = \int\sin x \cdot \sin (\cos x) dx

Let \cos x=\mathrm{t}

\therefore-\sin x d x=t

I= \int \sin x \cdot \sin (\cos x) d x=-\int \sin t d t

=-[-\cos t]+C

=\cos t+C

=\cos (\cos x)+C

where \mathrm{C} is an arbitrary constant.

Question 5:  \sin (a x+b) \cos (a x+b)

Solution: The given function can be rewritten as

\sin (a x+b) \cos (a x+b)=\frac{2 \sin (a x+b) \cos (a x+b)}{2}=\frac{\sin 2(a x+b)}{2}

Let I = \int \frac{\sin 2(a x+b)}{2}

\text { let } 2(a x+b)=t

\therefore 2 a d x=d t

\Rightarrow \int \frac{\sin 2(a x+b)}{2} d x=\frac{1}{2} \int \frac{\sin t d t}{2 a}

=\frac{1}{4 a}[-\cos t]+C

=\frac{-1}{4 a} \cos 2(a x+b)+C

Question 6: \sqrt{a x+b}

Solution: Let I= \int \sqrt{a x+b } dx

a x+b=t

\Rightarrow a d x=d t

\Rightarrow d x=\frac{1}{a} d t

\Rightarrow \int(a x+b)^{\frac{1}{2}} d x=\frac{1}{a} \int t^{\frac{1}{2}} d t

I=\frac{1}{a}\left(\frac{t^{3/2}}{\frac{3}{2}}\right)+C

=\frac{2}{3 a}(a x+b)^{\frac{3}{2}}+C

Question 7:  x \sqrt{x+2}

Solution : Let I =\int x \sqrt{x+2} \ dx

Let \mathrm{x}+2=\mathrm{t}

\therefore d x=d t

\Rightarrow \int x \sqrt{x+2} d x=\int(t-2) \sqrt{t d t}

=\int\left(t^{\frac{3}{2}}-2 t^{\frac{1}{2}}\right) d t

=\int t^{\frac{3}{2}} d t-2 \int t^{\frac{1}{2}} d t

=\frac{t^{\frac{5}{2}}}{\frac{5}{2}}-2\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C

=\frac{2}{5} t^{\frac{5}{2}}-\frac{4}{3} t^{\frac{3}{2}}+C

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

Question 8: x \sqrt{1+2 x^2}

Solution: Let I=\int x \sqrt{1+2 x^2}\ dx

Let 1+2 \mathrm{x}^2=\mathrm{t}

\therefore 4 \mathrm{xdx}=\mathrm{dt}

\Rightarrow dx = \frac{1}{4x}dt

\Rightarrow \int x \sqrt{1+2 x^2} d x=\int \frac{\sqrt{t}}{4} d t

=\frac{1}{4} \int t^{\frac{1}{2}} d t

=\frac{1}{4}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C

=\frac{1}{6}\left(1+2 x^2\right)^{\frac{3}{2}}+C

Question 9: Integrate (4 x+2) \sqrt{x^2+x+1}

Solution : Let I = \int (4 x+2) \sqrt{x^2+x+1}\ dx

Let x^2+x+1=t

\therefore(2 x+1) d x=d t

\Rightarrow dx = \frac{1}{2x+1}dt

I=\int 2(2 x+1) \sqrt{x^2+x+1} d x

=\int 2 \sqrt{t} d t

=2 \int \sqrt{t d t}

=2\left(\frac{t^{3/2}}{\frac{3}{2}}\right)+C

=\frac{4}{3}\left(x^2+x+1\right)^{\frac{3}{2}}+C

Question 10: \frac{1}{x-\sqrt{x}}

Solution: The given function can be rewritten as

\frac{1}{x-\sqrt{x}}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}

Let I = \int \frac{1}{\sqrt{x}(\sqrt{x}-1)}\ dx

\text { Let }(\sqrt{\mathrm{x}}-1)=\mathrm{t}

\therefore \frac{1}{2 \sqrt{x}} \mathrm{dx}=\mathrm{dt}

\Rightarrow dx = 2\sqrt{x}\ dt

\Rightarrow \int \frac{1}{\sqrt{x}(\sqrt{x}-1)} \mathrm{dx}=\int \frac{2}{\mathrm{t}} \mathrm{dt}

=2 \log |\mathrm{t}|+\mathrm{C}

=2 \log |\sqrt{x}-1|+C

Question 11:  \frac{x}{\sqrt{x+4}}, x>0

Solution : Let I = \int \frac{x}{\sqrt{x+4}}\ dx

Let x+4=t

\therefore d x=d t

\int \frac{x}{\sqrt{x+4}} d x=\int \frac{(t-4)}{\sqrt{t}} d t

=\int\left(\sqrt{t}-\frac{4}{\sqrt{t}}\right) d t

=\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)+C

=\frac{2}{3}(t)^{\frac{3}{2}}-8(t)^{\frac{1}{2}}+C

=\frac{2}{3} t \cdot t^{1/2}-8 t^{1/2}+C

=\frac{2}{3} t^{\frac{1}{2}}(t-12)+C

=\frac{2}{3}(x+4)^{\frac{1}{2}}(x+4-12)+C

=\frac{2}{3} \sqrt{x+4}(x-8)+C

Question 12: \left(x^3-1\right)^{\frac{1}{3}} x^5

Solution: Let I = \int \left(x^3-1\right)^{\frac{1}{3}} x^5\ dx

Let x^3-1=t

\therefore 3 \mathrm{x}^2 \mathrm{dx}=\mathrm{dt}

\Rightarrow dx = \frac{1}{3x^2} dt

I= \int\left(x^3-1\right)^{\frac{1}{3}} x^5 d x

=\int\left(x^3-1\right)^{\frac{1}{3}} x^3 x^2 d x

=\int t^{\frac{1}{3}}(t+1) \frac{d t}{3}

=\frac{1}{3} \int\left(t^{\frac{4}{3}}+t^{\frac{1}{3}}\right) d t

=\frac{1}{3}\left[\frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\right]+C

=\frac{1}{3}\left[\frac{3}{7} t^{\frac{7}{3}}+\frac{3}{4} t^{\frac{4}{3}}\right]+C

=\frac{1}{7}\left(x^3-1\right)^{\frac{7}{3}}+\frac{1}{4}\left(x^3-1\right)^{\frac{4}{3}}+C

Question 13: \frac{x^2}{\left(2+3 x^2\right)^2}

Solution: Let I= \int \frac{x^2}{\left(2+3 x^2\right)^2}\ dx

Let 2+3 x^3=t

\therefore 9 x^2 d x=d t

\Rightarrow \int \frac{x^2}{\left(2+3 x^3\right)^3} d x=\frac{1}{9} \int \frac{d t}{(t)^3}

=\frac{1}{9}\left|\frac{t^{-2}}{-2}\right|+C

=\frac{-1}{18}\left(\frac{1}{t^2}\right)+C

=\frac{-1}{18\left(2+3 x^3\right)^2}+C

Question 14: Integrate \frac{1}{x(\log x)^m}, x>0

Solution: Let I = \int \frac{1}{x(\log x)^m}\ dx

Let \log x=t

\frac{1}{x} d x=d t

\Rightarrow \int \frac{1}{x(\log x)^m} d x=\int \frac{d t}{(t)^m}

=\frac{t^{-m+1}}{-m+1}+C

=\frac{(\log x)^{1-m}}{(1-m)}+C

Question 15: Integrate \frac{x}{9-4 x^2}

Solution: Let I=\int \frac{x}{9-4 x^2}dx

Let 9-4 \mathrm{x}^2=\mathrm{t}

\therefore-8 x d x=d t

\Rightarrow dx =\frac{dt}{-8x}

I= \int \frac{x}{9-4 x^2} d x=\frac{-1}{8} \int \frac{1}{t} d t

=\frac{-1}{8} \log |t|+C

=\frac{-1}{8} \log \left|9-4 x^2\right|+C

Question 16: Integrate e^{2 x+3}

Solution: Let I=\int e^{2 x+3} dx

Let 2 \mathrm{x}+3=\mathrm{t}

\therefore 2 d x=d t

\Rightarrow \int e^{2 x+3} d x=\frac{1}{2} \int e^t d t

=\frac{1}{2}\left(e^t\right)+C

=\frac{1}{2}\left(e^{2 x+3}\right)+C

Question 17: Integrate \frac{x}{\mathrm{e}^{x^2}}

Solution: Let I=\int \frac{x}{\mathrm{e}^{x^2}} dx

Let x^2=t

\therefore 2 x d x=d t

\Rightarrow \int \frac{x}{e^{x^2}} d x=\frac{1}{2} \int \frac{1}{e^t} d t

=\frac{1}{2} \int e^{-t} d t

=\frac{1}{2}\left(\frac{e^{-t}}{-1}\right)+C

=-\frac{1}{2} e^{-x^2}+C

=\frac{-1}{2 e^{x^2}}+C

Question 18:  \frac{e^{\tan^{-1}x}}{1+x^2}

Solution: Let I=\int\frac{e^{\tan^{-1}x}}{1+x^2}dx

Let \tan ^{-1} \mathrm{x}=\mathrm{t}

\therefore \frac{1}{1+\mathrm{x}^2} \mathrm{dx}=\mathrm{dt}

\Rightarrow dx=(1+x^2)dt

\Rightarrow \int \frac{e^{\tan^{-1}x}}{1+\mathrm{x}^2} \mathrm{dx}=\int \mathrm{e}^t d t

=\mathrm{e}^t+\mathrm{C}

=\mathrm{e}^{\tan^{-1}x}+\mathrm{C}

Question 19:  \frac{e^{2 x}-1}{e^{2 x}+1}

Solution: Let I= \int \frac{e^{2 x}-1}{e^{2 x}+1} dx

Dividing the given function’s numerator and denominator by \mathrm{e}^{\mathrm{x}}, we obtain.

I=\int \frac{\frac{\left(e^{2 x}-1\right)}{e^x}}{\frac{\left(e^{2 x}+1\right)}{e^x}}dx

=\int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx

\text { Let } \mathrm{e}^x+\mathrm{e}^{-x}=\mathrm{t}

\left(e^x-e^{-x}\right) d x=d t

\Rightarrow \int \frac{e^{2 x}-1}{e^{2 x}+1} d x=\int \frac{e^x-e^{-x}}{e^x+e^{-x}} d x

=\int \frac{d t}{t}

=\log |\mathrm{t}|+\mathrm{C}

=\log \left|e^x-e^{-x}\right|+C

Question 20: Integrate \frac{\mathrm{e}^{2 x}-\mathrm{e}^{-2 x}}{\mathrm{e}^{2 x}+\mathrm{e}^{-2 x}}

Solution: Let I=\int \frac{\mathrm{e}^{2 x}-\mathrm{e}^{-2 x}}{\mathrm{e}^{2 x}+\mathrm{e}^{-2 x}} dx

Let \mathrm{e}^{2 x}+\mathrm{e}^{-2 x}=\mathrm{t}

\Rightarrow 2 e^{2 x}-2 e^{-2 x} d x=d t

\Rightarrow 2\left(e^{2 x}-e^{-2 x}\right) d x=d t

I= \int\left(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\right) d x=\int \frac{d t}{2 t}

=\frac{1}{2} \int \frac{1}{t} dt

=\frac{1}{2} \log |t|+C

=\frac{1}{2} \log \left|e^{2 x}+e^{-2 x}\right|+C

Question 21:\tan ^2(2 x-3)

Solution : Let I=\int \tan ^2(2 x-3)dx

=\int (\sec ^2(2 x-3)-1)dx

Let 2 \mathrm{x}-3=\mathrm{t}

\therefore 2 \mathrm{dx}=\mathrm{dt}

\Rightarrow \int \tan ^2(2 x-3) d x=\int\left\lfloor\sec ^2(2 x-3)-1\right] d x

I =\int \sec^2(2x-3)dx-\int 1dx

I=\frac{1}{2} \int\left(\sec ^2 t\right) d t-\int 1 d x

=\frac{1}{2} \int \sec ^2 t d t-\int 1 d x

=\frac{1}{2} \tan t-x+C

=\frac{1}{2} \tan (2 x-3)-x+C

Question 22:  \sec ^2(7-4 x)

Solution: Let I=\int \sec ^2(7-4 x) dx

Let 7-4 \mathrm{x}=\mathrm{t}

\therefore-4 \mathrm{dx}=\mathrm{dt}

\Rightarrow dx = -\frac{1}{4}dt

I= \int \sec ^2(7-4 x) d x=\frac{-1}{4} \int \sec ^2 \mathrm{tdt}

=\frac{-1}{4}(\tan t)+C

=\frac{-1}{4} \tan (7-4 x)+C

Question 23: Integrate \frac{\sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}

Solution: I=\int \frac{\sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}} dx

Let \sin ^{-1} x=t

\frac{1}{\sqrt{1-x^2}} d x=d t

I=\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x=\int t d t

=\frac{t^2}{2}+C

=\frac{\left(\sin ^{-1} x\right)^2}{2}+C

Question24:Integrate \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}

Solution: Let I=\int \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}dx

\Rightarrow I = \int \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}dx

=\int \frac{2 \cos x-3 \sin x}{2(3 \cos x+2 \sin x)}dx

\text { Let } 3 \cos x+2 \sin x=t

(-3 \sin x+2 \cos x) d x=d t

I=\int \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x} d x=\int \frac{d t}{2 t}

=\frac{1}{2} \int \frac{1}{t} d t

=\frac{1}{2} \log |t|+C

=\frac{1}{2} \log |2 \sin x+3 \cos x|+C

Question 25: Integrate \frac{1}{\cos ^2 x(1-\tan x)^2}

solution: Let I = \int \frac{1}{\cos ^2 x(1-\tan x)^2}dx

I = \int \frac{1}{\cos ^2 x(1-\tan x)^2}dx

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

\text { Let }(1-\tan x)=t

\Rightarrow -\sec ^2 x d x=d t

\Rightarrow dx = -\frac{1}{\sec^2x}

I = \int \frac{\sec ^2}{(1-\tan x)^2} d x=\int \frac{-d t}{t^2}

=-\int t^{-2} d t

=+\frac{1}{t}+C

=\frac{1}{(1-\tan x)}+C

Questio 26: integrate \frac{\cos \sqrt{x}}{\sqrt{x}}

Solution: Let I=\int \frac{\cos \sqrt{x}}{\sqrt{x}}dx

let \sqrt{x}=t

\frac{1}{2 \sqrt{x}} d x=d t

\Rightarrow dx = 2\sqrt{x}dt

I= \int \frac{\cos \sqrt{x}}{\sqrt{x}} d x

=2 \int \cos t d t=2 \sin t+C

=2 \sin \sqrt{x}+C

Question 27:  \sqrt{\sin 2 x} \cos 2 x

Solution: Let I = \int \sqrt{\sin 2 x} \cos 2 x\ dx

Let \sin 2 \mathrm{x}=\mathrm{t}

So, 2 \cos 2 x d x=d t

\Rightarrow dx = \frac{1}{2\cos 2x}dt

I= \int \sqrt{\sin 2 x} \cos 2 x d x

=\frac{1}{2} \int \sqrt{\mathrm{t} }\ dt

=\frac{1}{2}\left(\frac{\mathrm{t}^{3/2}}{\frac{3}{2}}\right)+\mathrm{C}

=\frac{1}{3} \mathrm{t}^{\frac{3}{2}}+\mathrm{C}

=\frac{1}{3}(\sin 2 x)^{\frac{3}{2}}+\mathrm{C}

Question 28: \frac{\cos x}{\sqrt{1+\sin x}}

Solution: Let I =\int \frac{\cos x}{\sqrt{1+\sin x}}dx

Let 1+\sin x=t

\therefore \cos x d x=d t

\Rightarrow dx = \frac{1}{\cos x}dt

I= \int \frac{\cos x}{\sqrt{1+\sin x}} d x

=\int \frac{d t}{\sqrt{t}}

=\frac{t^{1/2}}{\frac{1}{2}}+C

= 2 \sqrt{t}+C

= 2 \sqrt{1+\sin x}+C

Question 29: Integrate \cot x \log \sin x

Solution: Let I=\int \cot x \log \sin xdx

Let logsin x=t

\Rightarrow \frac{1}{\sin x} \cdot \cos x d x=d t

\therefore \cot x d x=d t

\Rightarrow dx =\frac{1}{\cot x}dt

I= \int \cot x \log \sin x d x=\int t d t

=\frac{t^2}{2}+C

=\frac{1}{2}(\log \sin x)^2+C

Question 30:  \frac{\sin x}{1+\cos x}

Solution: I= \int \frac{\sin x}{1+\cos x} d x

Let 1+\cos x = t

\Rightarrow -\sin x dx = dt

\Rightarrow dx =-\frac{1}{\sin x}dt

I=\int-\frac{\mathrm{dt}}{\mathrm{t}}

=-\log |\mathrm{t}|+\mathrm{C}

=-\log |1+\cos x|+C

Question 31: \frac{\sin x}{(1+\cos x)^2}

Solution:  Let I=\int \frac{\sin x}{(1+\cos x)^2}dx

Let 1+\cos x=t

\therefore-\sin x d x=d t

\Rightarrow dx=-\frac{1}{\sin x}dt

I = \int \frac{\sin x}{(1+\cos x)^2} d x=\int-\frac{d t}{t^2}

=-\int \mathrm{t}^{-2} \mathrm{dt}

=\frac{1}{\mathrm{t}}+\mathrm{C}

=\frac{1}{1+\cos \mathrm{x}}+\mathrm{C}

Question 32: \frac{1}{1+\cot x}

Solution: Let \mathrm{I}=\int \frac{1}{1+\cot x} \mathrm{dx}

=\int \frac{1}{1+\frac{\cos x}{\sin x}} d x

=\int \frac{\sin x}{\sin x+\cos x} d x

=\frac{1}{2} \int \frac{2 \sin x}{\sin x+\cos x} d x

=\frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{(\sin x+\cos x)} d x

=\frac{1}{2} \int 1 d x+\frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} d x

I =\frac{1}{2}(x)+\frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} d x

Let \sin x+\cos x=t \Rightarrow(\cos x-\sin x) d x=d t

\therefore I=\frac{x}{2}+\frac{1}{2} \int \frac{-(d t)}{t}

=\frac{x}{2}-\frac{1}{2} \log |t|+C

=\frac{x}{2}-\frac{1}{2} \log |\sin x+\cos x|+C

Question 33: \frac{1}{1-\tan x}

Solution : Let \mathrm{I}=\int \frac{1}{1-\tan \mathrm{x}} \mathrm{dx}

=\int \frac{1}{1-\frac{\sin x}{\cos x}} d x

=\int \frac{\cos x}{\cos x-\sin x} d x

=\frac{1}{2} \int \frac{2 \cos x}{\cos x-\sin x} d x

=\frac{1}{2} \int \frac{(\cos x-\sin x)+(\cos x+\sin x)}{(\cos x-\sin x)} d x

=\frac{1}{2} \int 1 d x+\frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} d x

=\frac{x}{2}+\frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} d x

Put \cos x-\sin x=t \Rightarrow(-\sin x-\cos x) d x=d t

\therefore I=\frac{x}{2}+\frac{1}{2} \int \frac{-(d t)}{t}

=\frac{x}{2}-\frac{1}{2} \log |t|+C

=\frac{x}{2}-\frac{1}{2} \log |\cos x-\sin x|+C

Question 34: \frac{\sqrt{\tan x}}{\sin x \cos x}

Solution: Let I=\int \frac{\sqrt{\tan x}}{\sin x \cos x} d x

=\int \frac{\sqrt{\tan x} \times \cos x}{\sin x \cos x \times \cos x} d x

=\int \frac{\sqrt{\tan x}}{\tan x \cos ^2 x} d x

=\int \frac{\sec ^2 x d x}{\sqrt{\tan x}}

Let \tan x=t \Rightarrow \sec ^2 x d x=d t

\Rightarrow dx =\frac{1}{\sec^2x}dt

\therefore I=\int \frac{d t}{\sqrt{t}}

=2 \sqrt{t}+C

=2 \sqrt{\tan x+C}

Question 35: Integrate \frac{(1+\log x)^2}{x}

Solution: Let I =\int \frac{(1+\log x)^2}{x}dx

Let 1+\log x=t

\therefore \frac{1}{x} d x=d t

I= \int \frac{(1+\log x)^2}{x} d x=\int t^2 d t

=\frac{t^3}{3}+C

=\frac{(1+\log x)^3}{3}+C

Question 36:  \frac{(x+1)(x+\log x)^2}{x}

Solution: Let I = \int \frac{(x+1)(x+\log x)^2}{x}dx

\Rightarrow I=\int \left(1+\frac{1}{x}\right)(x+\log x)^2dx

\text { Let }(x+\log x)=t

\therefore\left(1+\frac{1}{x}\right) d x=d t

I = \int\left(1+\frac{1}{x}\right)(x+\log x)^2 d x

=\int t^2 d t

=\frac{t^3}{3}+C

=\frac{1}{3}(x+\log x)^3+C

Question 37: \frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{1+x^8}

Solution: Let I = \int\frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{1+x^8}dx

Let \tan ^{-1} x^4=t

\frac{1}{1+\mathrm{x}^4} 4x^3\mathrm{dx}=\mathrm{dt}

\Rightarrow dx =\frac{1+x^4}{4x^3}dt

I = \int \frac{x^3 \sin \left(\tan ^{-1} x^4\right) d x}{1+x^8}=\frac{1}{4} \int \text { sintdt }

=\frac{1}{4}(-\cos t)+C

=-\frac{1}{4} \cos \left(\tan ^{-1} x^4\right)+C

=\frac{-1}{4} \cos \left(\tan ^{-1} x^4\right)+C

Question 38: \int \frac{10 x^9+10^x \log _6 10}{x^{10}+10^x} d x equals

a) 10^x-x^{10}+C

b) 10^x+x^{10}+C

c) \left(10^x-x^{10}\right)^{-1}+C

d) \log \left(10^x+x^{10}\right)+C

Solution : Hence, the correct Answer is D

Let I=\int \frac{10 x^9+10^x \log _6 10}{x^{10}+10^x} d x

Let x^{10}+10^x=t

\therefore\left(10 x^9+10^x \log _e 10\right) d x=\int \frac{d t}{t}

\Rightarrow \int \frac{10 x^9+10^x \log _c 10}{x^{10}+10 x} d x=\int \frac{d t}{t}

=\log t+C

=\log \left(10^x+x^{10}\right)+C

Hence, the correct Answer is D.

Question 39: \int \frac{d x}{\sin ^2 x \cos ^2 x} equals

a) \tan x+\cot x+C

b) \tan x-\cot x+C

c) \tan x \cot x+c

d) \tan x-\cot 2 x+c

Solution: Hence, the correct Answer is B.

Let I=\int \frac{d x}{\sin ^2 x \cos ^2 x}

=\int \frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x} d x

=\int \frac{\sin ^2 x}{\sin ^2 x \cos ^2 x} d x+\int \frac{\cos ^2 x}{\sin ^2 x \cos ^2 x} d x

=\int \sec ^2 x d x+\int \operatorname{cosec}^2 d x

=\tan x-\cot x+C

Hence, the correct Answer is B.

 


Find an anti derivative (or integral) of the following functions by the method of inspection.(Ex 7.1 integration ncert solution maths class 12)

 

integration multiple choice question

 

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