CBSE Questions for Class 12 Commerce Maths Integrals Quiz 3 - MCQExams.com

Consider, $$I=\displaystyle \int_{0}^{1}\sin \left (2 tan^{-1}\sqrt{\dfrac{1-x}{1+x}} \right ) dx$$

let, $$x=  \cos \theta$$ $$\Rightarrow$$ $$dx=-\sin \theta  d \theta$$

$$\Rightarrow$$ $$x\rightarrow 0 \Rightarrow  \cos \theta \rightarrow \dfrac{\pi}{2}$$ and $$x\rightarrow 1 \Rightarrow  \cos \theta \rightarrow 0$$

$$\Rightarrow$$ $$I=\displaystyle \int_{\frac{\pi}{2}}^{0} \sin ( 2 tan^{-1}(tan(\dfrac{\theta}{2})) (-\sin \theta)  d\theta$$

$$\Rightarrow$$ $$I=\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^2 \theta  d \theta = \displaystyle \int_{0}^{\frac{\pi}{2}}\left [ \dfrac{1-\cos 2 \theta }{2} \right ] d\theta$$

$$\Rightarrow$$ $$I=\displaystyle \int_{0}^{\frac{\pi}{2}}\dfrac{1}{2}d  \theta - \dfrac{1}{2}\displaystyle \int_{0}^{\frac{\pi}{2}}\cos 2 \theta d \theta$$

$$\Rightarrow$$ $$I=\left[\left ( \dfrac{\pi}{4}-0 \right )+\dfrac{1}{2}  \sin 2 \theta \right]_{0}^{\frac{\pi}{2}}$$

$$\Rightarrow$$ $$I=\dfrac{\pi}{4}$$

$$\int_{0}^{\dfrac{\pi}{2}}\left ( 2  tan \dfrac{x}{2} + x  sec^2  \dfrac{x}{2} \right )  dx$$
$$=2 \int_{0}^{\dfrac{\pi}{2}}\left ( tan \dfrac{x}{2} + \dfrac{1}{2} x sec^2  \dfrac{x}{2} \right )  dx$$
$$=2 \int_{0}^{\dfrac{\pi}{2}}d\left ( x  tan \dfrac{x}{2} \right )$$
$$=2  \left ( x  tan \dfrac{x}{2} \right ) \int_{0}^{\dfrac{\pi}{2}}$$
$$=2 \left [ \left ( \dfrac{\pi}{2} 1 \right ) - 0 \right ]$$
$$=\pi$$

Consider, $$\displaystyle I=\int_{-\pi/4}^{\pi /4}log (cos  x +  sin  x)dx$$

$$I=\displaystyle \int_{-\dfrac{\pi}{4}}^{\dfrac{\pi}{4}}log \left [ \sqrt{2} sin \left ( \dfrac{\pi}{4}+ x \right ) \right ]  dx$$                

let, $$\dfrac{\pi}{4}+x=t$$ $$\Rightarrow$$ $$dx=dt$$

So, $$x \rightarrow \dfrac{-\pi}{4} \Rightarrow t \rightarrow 0$$ and $$x \rightarrow \dfrac{\pi}{4} \Rightarrow t \rightarrow \dfrac{\pi}{2}$$

$$\Rightarrow$$ $$\displaystyle I=\int_{0}^{\dfrac{\pi}{2}} \log (\sqrt{2} \sin  t)dt$$

$$\Rightarrow$$ $$\displaystyle I=\int_{0}^{\dfrac{\pi}{2}} \log (\sqrt{2}) + \int_{0}^{\dfrac{\pi}{2}} \log (\sin  t)dt$$

$$\Rightarrow$$ $$\displaystyle I=\dfrac{\pi}{2}\log (\sqrt{2})-\dfrac{\pi}{2} \log (2)$$

$$\Rightarrow$$ $$\displaystyle I=\dfrac{\pi}{4}\log  2 -\dfrac{\pi}{2} \log (2)$$

$$\Rightarrow$$ $$\displaystyle I=-\dfrac{\pi}{4} \log  2$$

$$\displaystyle \int_{0}^{a}\sqrt{\dfrac{a+x}{a-x}}dx$$

$$x=a  cos  \theta =dx -a  sin  \theta  d  \theta$$

$$=\displaystyle \int_{\dfrac{\pi}{2}}^{a}\sqrt{\dfrac{a+ a  cos  \theta}{a-  a  cos  \theta}}(-a  sin  \theta)d  \theta$$

$$=a\displaystyle \int_{0}^{\dfrac{\pi}{2}} \sqrt{\dfrac{1+ cos  \theta}{1- cos  \theta}}sin  \theta  d  \theta$$

$$=a\displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{cos \dfrac{\theta}{2}}{sin \dfrac{\theta}{2}}\cdot  2 sin \dfrac{\theta}{2} d\theta$$

$$=a\displaystyle \int_{0}^{\dfrac{\pi}{2}}2  cos^2 \dfrac{\theta}{2} d \theta = a \displaystyle \int_{0}^{\dfrac{\pi}{2}}(cos \theta -1) d \theta$$

$$=a \left ( sin \theta\right )_{0}^{\frac{\pi}{2}} - (\theta)_{0}^{\frac{\pi}{2}}$$

$$=a[1-0]-\left (\dfrac{\pi}{2} \right)$$

$$=a\left (\dfrac{\pi}{2}+1 \right )$$

$$=\dfrac{a}{2}(\pi+2)$$

$$\int_{0}^{1} \dfrac{dx}{(x+\sqrt{x})}$$
$$x=t^2$$
$$\int_{0}^{1}\dfrac{2t  dt}{t^2+ t}=2\int_{0}^{1}\dfrac{dt}{t+1}=2 log (t+1)\int_{0}^{1}$$
$$=2  log (2)-2log (1)$$
$$=2  log  2$$

$$\int_{0}^{k}\dfrac{dx}{2+ 8x^2}=\dfrac{\pi}{16}$$
$$\dfrac{1}{2}\int_{0}^{k}\dfrac{dx}{1+(2x)^2}=\dfrac{\pi}{16}$$
$$\dfrac{1}{2}(\dfrac{1}{2})tan^{-1}(2x)\int_{0}^{k}=\dfrac{\pi}{16}$$
$$\dfrac{1}{4}tan^{-1}(2k)=\dfrac{\pi}{16}$$
$$tan^{-1}(2k)=\dfrac{\pi}{4}$$
$$2k=1$$
$$k=\dfrac{1}{2}$$

$$I_{n}=\displaystyle \int_{^{\pi}/_{4}}^{^{\pi}/_{2}}(cot\ \theta)^{n}$$

$$I_{n}+I_{n+2}=\displaystyle \int_{^{\pi}/_{4}}^{^{\pi}/_{2}}[(cot\ \theta)^{n}+(cot\ \theta)^{n+2}]d\ \theta$$

$$=\displaystyle \int_{^{\pi}/_{4}}^{^{\pi}/_{2}}(cot\ \theta)^{n}cosec^{2}d\ \theta$$

$$=\displaystyle \int_{^{-\pi}/_{4}}^{^{\pi}/_{2}}(cot\ \theta)^{n}d(cot\ \theta)$$

$$-\dfrac{(cot\ \theta)^{n+1}}{n+1}|_{^{-\pi}/_{4}}^{^{\pi}/_{2}}$$

$$-\left [ 0-\dfrac{1}{n+1} \right ]$$

$$=\dfrac{1}{n+1}$$

$$\displaystyle \int_{1}^{\sqrt[7]{2}}\dfrac{1}{x(2x^{7}+1)}dx=\displaystyle \int_{1}^{\sqrt[7]{2}}\dfrac{1}{x^{8}(2+1/x^{7})dx}$$

$$1/x^{7}= 1\Rightarrow\dfrac{-7}{x^{8}}dx=dt$$

$$\Rightarrow \dfrac{dx}{x^{8}}-\dfrac{dx}{7}$$

$$\Rightarrow \dfrac{1}{-7}\displaystyle \int_{1}^{1/2}\dfrac{dt}{(2+t)}=\dfrac{1}{7}\displaystyle \int_{1/2}^{1}\left [ \log(2+t) \right ]_{1/2}^{1}$$

$$=\dfrac{1}{7} \left [ \log(3)-\log\dfrac{5}{2} \right ]$$

$$=\dfrac{1}{7}  \log \dfrac{6}{5}$$

$$\displaystyle \dfrac{df(x)}{dx}=\displaystyle \dfrac{e^{sin  x}}{x} ,  x  >  0,  \int_1^4  \dfrac{3e^{sin x^3}}{x}dx=f(k)-f(1)$$
substitute  $$x^3=p$$
$$dp=3x^2  dx$$
limit of p will be rom $$1   \rightarrow  64$$  as of  x  was  $$1  \rightarrow  4$$
So, $$\displaystyle \int_1^{64} \dfrac{3e^{sin  p}}{x} \cdot  \dfrac{dp}{3x^2}=\int_1^{64} \dfrac{e^{sin  p}}{p}dp=\int_1^{64}df(p)=f(64)-f(1)$$