CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 3 - MCQExams.com

$$x=t^2$$
$$dx=2t   dt$$
$$=\int_{0}^{1}\left ( \dfrac{t^2}{1+t} \right )  2t \cdot  dt$$
$$=2\int_{0}^{1}\dfrac{t^3}{1+t}dt =2\int_{0}^{1}\dfrac{t^3+1}{t+1}dt -2\int_{0}^{1}\dfrac{1}{t+1}dt$$
$$=2\left [ \dfrac{t^3}{3}\dfrac{t^2}{2}+t \right] \int_{0}^{1} -2 log (1+1) \int_{0}^{1}$$
$$=2\left [ \dfrac{1}{3} -\dfrac{1}{2} +1 \right ] -2 log (2)$$
$$=2 \left [ \dfrac{2-3+6}{6} \right ] -2  log (2)$$
$$=\dfrac{5}{3}-log  4$$

$$\int_{0}^{\pi}\dfrac{dx}{3+2  sin  x+  cos  x}$$
$$=\int_{0}^{\pi}\dfrac{dx(1+tan^2 \dfrac{x}{2})}{3(1+tan^2 \dfrac{x}{2}) + 2 (2 tan \dfrac{x}{2}) + (1- tan^2 \dfrac{x}{2})}$$
$$=\int_{0}^{\pi}\dfrac{sec^2\dfrac{x}{2} dx}{2 tan^2 \dfrac{x}{2}+ 4 tan \dfrac{x}{2}+4}$$
$$=\int_{0}^{\pi}\dfrac{2 d (tan^2 \dfrac{x}{2})}{(\sqrt{2}tan \dfrac{x}{2}+\sqrt{2})^2+(\sqrt{2})^2}$$
$$=\int_{0}^{\pi}\dfrac{d (tan^2 \dfrac{x}{2})}{(\sqrt{2}tan \dfrac{x}{2}+\sqrt{2})^2+(\sqrt{2})^2}$$
$$=\int_{0}^{\pi}\dfrac{d(tan^2 \dfrac{x}{2})}{(tan \dfrac{x}{2}+1)^2} =tna^{-1}(tan \dfrac{x}{2}+1)\int_{0}^{\pi}$$
$$=\dfrac{\pi}{2} -\dfrac{\pi}{4}$$
$$=\dfrac{\pi}{4}$$

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

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 $$x \rightarrow \dfrac{\pi}{4} \Rightarrow t \rightarrow \dfrac{\pi}{2}$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$=\dfrac{1}{7}  \log \dfrac{6}{5}$$

$$\displaystyle \int_{0}^{\pi/2}\dfrac{sinx+cosx}{3+sin2x}dx.$$

$$\displaystyle \int_{0}^{\pi/2}\dfrac{sinx+cosx}{4-(sinx-cosx)^{2}}$$

$$dx= \displaystyle \int_{0}^{\pi/2}\dfrac{d(sinx-cosx)}{(2)^{2}-(sinx-cosx)^{2}}$$

$$=\dfrac{1}{4}\log\left | \dfrac{2+(sinx-cosx)}{2-(sinx-cosx)} \right |_{0}^{\pi}$$

$$=\dfrac{1}{4}\left ( \log\left | \dfrac{2+0}{2-0} \right | -\log\left|\dfrac{2-1}{2+1}\right | \right )$$

$$=\dfrac{1}{4} (-\log \dfrac{1}{3})$$

$$= \dfrac{1}{4} \log(3)$$

$$\int_{0}^{1}\left ( \dfrac{2^{2x+1}+5^{2x-1}}{10x} \right )dx$$
$$2\int_{0}^{1}\left ( \dfrac{4}{10} \right )^{xdx}-1/5\int_{0}^{1}\dfrac{25}{10}^{x}dx$$
$$=2\dfrac{(\dfrac{4}{10})^{x}}{log\dfrac{(4)}{10}}\int_{0}^{1}-\dfrac{\dfrac{1}{5}(\dfrac{25}{10})}{log(\dfrac{25}{10})}$$
$$=\dfrac{2\left [ (\dfrac{4}{10})-1 \right ]}{log(2/5)}-\dfrac{1/5\left [ (\dfrac{25}{10}-1) \right ]}{log(5/2)}$$
$$\dfrac{2(-3/5)}{l0g(2/5)}-\dfrac{1/5(3/2)}{log(5/2)}$$
$$=-3/5\left [ \dfrac{2}{log(2/5)}+\dfrac{1}{2log}(5/2) \right ].$$