Integrals Test - 17

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

$$\int \dfrac{1}{a^2+x^2}dx=\dfrac{1}{a^2}$$

$$\int \dfrac{d(\dfrac{x}{a})}{1+(\dfrac{x}{a})^2}=\dfrac{1}{a}\int \dfrac{d(\dfrac{x}{a})}{1+(\dfrac{x}{a})^2}$$

$$=\dfrac{1}{a}tan^{-1}(\dfrac{x}{a})+c$$

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

$$=\left [ \dfrac{1}{a}tan^{-1}\left ( \dfrac{a}{a} \right ) +c \right ] -(0+c)$$
$$=\dfrac{1}{a}tan^{-1}(1)$$

$$=\dfrac{1}{a}\cdot \dfrac{\pi}{4}$$

$$=\dfrac{\pi}{4a}$$

$$\displaystyle \int_{a}^{\frac{a}{2}}\dfrac{1}{\sqrt{a^2-x^2}}\cdot dx$$

$$\displaystyle \int \dfrac{1}{\sqrt{a^2-x^2}}\cdot dx=\dfrac{1}{a}\displaystyle \int \dfrac{1}{\sqrt{1-\left ( \dfrac{x}{a} \right )^2}}\cdot dx$$

$$=\displaystyle \int \dfrac{d\left ( \dfrac{x}{a} \right )}{\sqrt{1-\left ( \dfrac{x}{a} \right )^2}}$$

$$=\sin^{-1}(\dfrac{x}{a})+c$$

$$\displaystyle \int_{\frac{a}{2}}^{a}\dfrac{1}{\sqrt{a^2-x^2}}\cdot =\left [ \sin^{-1}(\dfrac{x}{a})+c \right ] _{\frac{a}{2}}^{a}$$ 

$$=(\sin^{-1}(\dfrac{a}{a})+c)-(\sin^{-1}(\dfrac{\frac{a}{2}}{a})+c)$$

$$=\left [ (\sin (1)+c ) -(\sin^{-1}(\dfrac{1}{2})+c) \right ]$$

$$=\dfrac{\pi}{2}-\dfrac{\pi}{6}$$

$$=\dfrac{\pi}{3}$$

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

$$\displaystyle \int \dfrac{x}{1+x^2}dx=\dfrac{1}{2}\displaystyle \int \dfrac{dx^2}{1+x^2}=\dfrac{1}{2} \log (1+x^2)+c$$

$$=\displaystyle \int_{0}^{1}\dfrac{x}{1+x^2}dx=\left [ \dfrac{1}{2} \log (1+x^2)+c \right ]_{0}^{1}$$

$$=\left ( \dfrac{1}{2} \log (2)+c \right ) - \left ( \dfrac{1}{2} \log (1)+c\right )$$

$$=\dfrac{1}{2} \log 2$$

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

Let $$\displaystyle I=\int_{0}^{1}\dfrac{x^3}{1+x^8} dx$$

Now, $$\displaystyle \int \dfrac{x^3}{1+x^8 }dx =\dfrac{1}{4}\int \dfrac{d(x^4)}{1+x^8}=\dfrac{1}{4}\tan^{-1}(x^4)+c$$

$$I=\displaystyle \int_{0}^{1}\dfrac{x^3}{1+x^8}dx=\left(\dfrac{1}{4}\tan^{-1}(x^4)+c\right)_{0}^{1}$$

$$=\left [\dfrac{1}{4} \tan^{-1}(1)+c\right ]-\left [\dfrac{1}{4} \tan^{-1}(0)+c\right ]$$

$$=\dfrac{1}{4} \tan^{-1}(1)$$

$$=\dfrac{1}{4} \cdot \dfrac{\pi}{4}=\dfrac{\pi}{16}$$

Hence, $$\displaystyle \int_{0}^{1}\dfrac{x^3}{1+x^8}dx =\dfrac{\pi}{16}$$

$$\int_{0}^{1}\dfrac{1}{1+x}dx=\ln(1+x)+c$$
$$\int_{0}^{1}\dfrac{1}{1+x}dx =\left [ \ln (1+x)+ c \right ]\int_{0}^{1}$$
$$=(\ln(2)+c)-(\ln(1)+c)$$
$$=\ln2$$
$$\int_{0}^{1}\dfrac{1}{1+x}dx=ln2$$

$$\displaystyle \int_{0}^{1}\dfrac{(\tan^{-1}x)^3}{1+x^2}dx$$

$$\displaystyle \int \dfrac{(\tan^{-1}x)^3}{1+x^2}dx = \int (\tan^{-1}x)^3 d (\tan^{-1}x)$$

$$=\dfrac{(\tan^{-1}x)^4}{4}+c$$

$$\displaystyle \int_{0}^{1}\dfrac{(\tan^{-1}x)^3}{1+x^2}\cdot dx=\left ( \dfrac{(\tan^{-1}x)^4}{4}+c \right )_{0}^{1}$$

$$=\left [ \dfrac{(\tan^{-1}1)^4}{4}+c \right ] - \left [ \dfrac{(\tan^{-1}0)^4}{4}+1 \right ] $$

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

$$=\dfrac{\pi^4}{4^{5}}=\dfrac{\pi^4}{1024}$$

Consider, $$I=\displaystyle \int_{0}^{\frac{\pi}{4}}\dfrac{e^{\tan x}}{\cos ^2  x}dx$$

$$\displaystyle \int e^{\tan x}\cdot \sec ^2 x  dx =\int e^{\tan x}\cdot d({\tan x})$$

$$=\int e^{\tan x} d(\tan  x )=e^{\tan x}+c$$

$$\therefore \displaystyle \int_{0}^{\frac{\pi}{4}} \dfrac{e^{\tan x}}{\cos ^2 x}dx=\left ( e^{\tan x}+ c \right )_{0}^{\frac{\pi}{4}}$$

$$=\left ( e^{\tan \frac{\pi}{4}}+c\right ) -\left ( e^{\tan 0}+c \right )$$

$$=(e+c)-(1+c)$$

$$=e-1$$

Hence, $$\displaystyle \int_{0}^{\frac{\pi}{4}}\dfrac{e^{\tan x}}{\cos ^2  x}dx=e-1$$

Let $$sin x = t$$
Thus $$cos x dx = dt$$