Integrals Test - 15

We need to find the value of $$\displaystyle \int_{-1}^{3} \left[ \tan^{-1}\left ( \dfrac{x}{x^2+1} \right) + \tan^{-1}\left ( \dfrac{x^2+1}{x}\right ) \right] dx$$

We know $$\tan^{-1}p+\cot^{-1}p=\dfrac{\pi}{2}$$

$$\Rightarrow   \tan^{-1}p+\tan^{-1}\dfrac{1}{p}=\dfrac{\pi}{2}$$

So, $$\tan^{-1}\left ( \dfrac{x}{x^2+1} \right )+\tan^{-1}\left ( \dfrac{x^2+1}{x} \right )=\dfrac{\pi}{2}$$

$$\displaystyle \int_{-1}^{3}\left [ \tan^{-1} \left ( \dfrac{x}{1+x^2}\right ) + \tan^{-1}\left ( \dfrac{x^2+1}{x}\right ) \right ]dx$$

$$=\displaystyle \int_{-1}^{3}\dfrac{\pi}{2}dx$$

$$=\dfrac{\pi}{2}\int_{-1}^{3}1\cdot dx$$

$$=\dfrac{\pi}{2}[3-(-1)]$$

$$=\dfrac{\pi}{2}\times 4=2\pi$$

$$\displaystyle \int_{\frac{1}{\sqrt{3}}}^{k}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{6}$$

$$\displaystyle \int_{\frac{1}{\sqrt{3}}}^{k}\dfrac{1}{1+x^2}dx=tan^{-1}x\displaystyle \vert _{\frac{1}{\sqrt{3}}}^{k}$$

$$=[tan^{-1}k-tan^{\dfrac{-1}{\sqrt{3}}}]$$

$$=tan^{-1}k-tan^{-1}\dfrac{1}{\sqrt{3}}=\dfrac{\pi}{6}$$

$$tan^{-1}\dfrac{1}{\sqrt{3}}=\dfrac{\pi}{6}$$ ---- equation (i)

$$tan^{-1}k=\dfrac{\pi}{6}+\dfrac{\pi}{6}$$ ---- equation (ii)

$$tan^{-1}k=\dfrac{\pi}{3}$$

$$k=\sqrt{3}$$

$$\displaystyle I=\int_{0}^{\dfrac{\pi}{4}}\dfrac{\sin ^9  x}{\cos ^{11}x}dx=\int_{0}^{\dfrac{\pi}{4}}\dfrac{\sin ^9  x}{\cos ^{9}x} \sec ^2  dx$$

$$\displaystyle =\int_{0}^{\dfrac{\pi}{4}} \tan ^9x d(\tan   x)$$

$$=\left[\dfrac{\tan ^{10}x}{10}+c \right]_{0}^{\dfrac{\pi}{4}}$$

$$=\left [ \left ( \dfrac{\tan ^{10}(\dfrac{\pi}{4})}{10}+c \right ) -\left ( \dfrac{\tan ^{10}(0)}{10} +c \right )\right ]$$

$$\tan \dfrac{\pi}{4}=1  ;  \tan   0=0$$

$$\Rightarrow I=\dfrac{1}{10}-\dfrac{0}{10}=\dfrac{1}{10}$$

So,  $$\displaystyle \int_{0}^{\dfrac{\pi}{4}}\dfrac{\sin ^9 x}{\cos ^{11}x}dx=\dfrac{1}{10}$$

$$\int_{0}^{\dfrac{\pi}{2}}\dfrac{cos  x}{1+sin  x}dx$$
we know that
$$\int_{0}^{\dfrac{\pi}{2}}\dfrac{cos  x}{1+sin  x}dx=\int_{0}^{\dfrac{\pi}{2}}\dfrac{d(sin  x)}{1+sin  x}$$
$$\int_{0}^{\dfrac{\pi}{2}}\dfrac{d (sin  x)}{1+sin  x}=log (1+sin  x)\int_{0}^{\dfrac{\pi}{2}}$$
$$=log (1+sin \dfrac{\pi}{2}) - log (1+sin  0)$$
$$=log (2)$$
$$=log 2$$

We need to find value of $$\displaystyle \int_{1}^{2}\dfrac{1}{x\sqrt{x^2-1}}dx$$
We know that

$$\displaystyle \int   \dfrac{1}{x\sqrt{x^2-1}}dx=\sec^{-1}x+c$$

$$\displaystyle \int_{1}^{2}\dfrac{1}{x\sqrt{x^2-1}}dx= \sec^{-1}2-\sec^{-1}1$$

$$\sec^{-1}2 =\dfrac{\pi}{3}$$

and $$\sec^{-1}1=0$$

So,  $$\sec^{-1}2-\sec^{-1}1=\dfrac{\pi}{3}$$

$$\therefore \displaystyle \int_{1}^{2}\dfrac{1}{x\sqrt{x^2-1}}dx=\dfrac{\pi}{3}$$

$$\int_{0}^{\infty}xe^{-x^2}dx=\dfrac{1}{2}\int_{0}^{\infty}e^{-x^2}dx^2$$
$$=\dfrac{1}{2}\int_{0}^{\infty}e^{-x^2}dx^2=\dfrac{1}{2}\left [ -\dfrac{1}{e^{x^2}}+c \right ] \int_{0}^{\infty}$$
$$=\dfrac{1}{2}\left [ (0+e)-(-1+e) \right ]$$
$$=\dfrac{1}{2}$$
$$\int_{0}^{\infty}xe^{-x^2}dx=\dfrac{1}{2}$$

$$\displaystyle \int_{0}^{1}\dfrac{x^2dx}{1+x^2}$$

$$\displaystyle \int_{0}^{1}1{dx}-\displaystyle \int_{0}^{1}\dfrac{1}{1+x^2}dx$$

$$=1-\left [ \tan^{-1}x \right ] \displaystyle \vert_{0}^{1}$$

$$=1 - \left [ \tan^{-1}  1 \right ]$$

$$=1 -\dfrac{\pi}{4}$$

Thus $$\displaystyle \int_{0}^{1}\dfrac{x^2}{1+x^2}dx=1-\dfrac{\pi}{4}$$

$$\int_{0}^{1}\dfrac{e^x}{1+e^{2x}}dx=\int_{0}^{1}\dfrac{de^x}{1+e^{2x}}$$
$$=tan^{-1}(e^x)\int_{0}^{1}$$
$$=tan^{-1}(e)-tan^{-1}(e^{\circ})$$
$$=tan^{-1}(e)-tan^{-1}(1)$$
$$=tan^{-1}(e)-\dfrac{\pi}{4}$$
So,   $$\int_{0}^{1}\dfrac{e^x}{1+e^{2x}}dx=tan^{-1}e-\dfrac{\pi}{4}$$