Integrals Test - 20

$$\displaystyle \int_{1}^{2}\sqrt{-x^2+3x-2}dx$$

$$=\displaystyle \int_{1}^{2}\sqrt{-x^2+3x-\dfrac{9}{4}+\dfrac{9}{4}-2} dx$$

$$=\displaystyle \int_{1}^{2}\sqrt{-\left (x-\dfrac{3}{2}\right)^2+\left (\dfrac{1}{2}\right)^2}dx=\int_{1}^{2}\sqrt{\left (\dfrac{1}{2}\right)^2-\left (x-\dfrac{3}{2}\right)^2}dx$$

$$=\left ( \dfrac{x}{2}\sqrt{(x-1)(2-x)+\dfrac{(\dfrac{1}{2})^2}{2}}\sin^{-1}\left ( \dfrac{x-\dfrac{3}{2}}{\dfrac{1}{2}} \right )+c \right )_{1}^{2}$$

$$=\left ( \dfrac{x}{2}\sqrt{(x-1)(2-x)}+\dfrac{1}{8}\sin^{-1}(2x-3)+c \right )_{1}^{2}$$

$$=\dfrac{1}{8}\sin^{-1}(1)-\dfrac{1}{8}\sin^{-1}(-1)$$

$$=\dfrac{1}{8}\left ( \dfrac{\pi}{2}+\dfrac{\pi}{8} \right )$$

$$=\dfrac{\pi}{8}$$

$$\displaystyle \int \sec^{2}x. \text{cosec}^{2}xdx$$

$$\displaystyle =\int \dfrac{1}{\cos^{2}x.\sin^{2}x}dx$$

$$\displaystyle =\int \dfrac{\cos^{2}x+\sin^{2}x}{\cos^{2}x.\sin^{2}x}dx$$

$$\displaystyle =\int \dfrac{\cos^{2}x}{\cos^{2}x\sin^{2}x}+\dfrac{\sin^{2}x}{\cos^{2}x.\sin^{2}x}dx$$

$$\displaystyle =\int \dfrac{1}{\sin^{2}x}+\dfrac{1}{\cos^{2}x}dx$$

$$\displaystyle =\int (\text{cosec}^{2}x+\sec^{2}x) dx$$

$$=-\cot x+\tan x +c$$

$$=\tan x-\cot x+c$$

$$\int_{0}^{\dfrac{\pi}{2}} sin ^4  x  cos^2  x  dx$$
$$\int_{0}^{\dfrac{\pi}{2}}sin ^m  x  cos ^x   dx$$
$$I_{m, n}=\left (\dfrac{m-1}{m+n} \right )\left (\dfrac{m-3}{m+n-2} \right )\left (\dfrac{m-5}{m+n-4} \right ) .....  I_0, n    or   I_1,,,,,$$
$$I_{4,2}=\left (\dfrac{4-1}{6} \right )\left (\dfrac{4-3}{4} \right )I_{0,2}$$
$$\int_{0}^{\dfrac{\pi}{2}}cos^2x   dx=\dfrac{1}{2}\times \dfrac{\pi}{2}$$
$$\dfrac{3}{6}\left (\dfrac{1}{4} \right )\times \dfrac{\pi}{4}=\dfrac{\pi}{32}$$

$$\int_{1}^{3}\dfrac{dx}{\sqrt{(x-1)(3-x)}}$$
$$\int_{1}^{3}\dfrac{dx}{\sqrt{-3+4x-x^2}}=\int_{1}^{3}\dfrac{dx}{\sqrt{1-(x-2)^2}}$$
$$=\int_{1}^{3}\dfrac{dx}{\sqrt{1-(x-2)^2}}=\left [ sin^{-1}\left (\dfrac{x-2}{1}\right )+c\right ]_1^3$$
$$=sin^{-1}(x-2)_{1}^{3}$$
$$=sin^{-1}(1)-sin^{-1}(-1)$$
$$=\dfrac{\pi}{2}-\left ( \dfrac{-\pi}{2}\right )$$
$$=\pi$$

$$\int_{0}^{\infty}\dfrac{dx}{(x+\sqrt{x^2+1})^5}=\int_{0}^{\infty}\dfrac{(x-\sqrt{x^2+1})^5}{-1}dx$$
$$=\int_{0}^{\infty}(\sqrt{x^2+1}-1)^5  dx$$
$$x=Sin  h  \theta$$
$$\int_{0}^{\infty}(cos  h  \theta-  sin  h  \theta)^5  cos  h  \theta  d\theta$$
$$\int_{0}^{\infty}(e^{-x})^5\left (\dfrac{e^x+e^{-x}}{2} \right ) d\theta$$
$$\int_{0}^{+\infty}\dfrac{e^{-4x}+e^{-6x}}{2}  dx=\dfrac{1}{2}\left [ \left (\dfrac{-1}{4}\right )[0-1]+\left (\dfrac{-1}{6} \right )[0-1] \right ]$$
$$=\dfrac{1}{2}\left ( \dfrac{1}{4}+\dfrac{1}{6}\right )=\dfrac{1}{2}\left (\dfrac{1+4}{24}\right )$$
$$=\dfrac{5}{24}$$

$$\int_{0}^{1}\sqrt{x(1-x)}dx=\int_{0}^{1}\sqrt{x-x^2}dx$$
$$\int_{0}^{1}\sqrt{(\frac{1}{2})-(x-\frac{1}{2})^2}dx$$
$$\int\sqrt{(\frac{1}{2})^2-(x-\frac{1}{2})^2}dx=\frac{x}{2}\sqrt{(\frac{1}{2})^2-(x-\frac{1}{2})^2}+\frac{(\frac{1}{2})^2}{2}sin^{-1}\left ( \frac{(x-\frac{1}{2})}{\frac{1}{2}} \right)+c$$
$$=\frac{x}{2}\sqrt{x-x^2}+\frac{1}{8}sin^{-1}(2x-1)+c$$
$$\int_{0}^{6}\sqrt{x(1-x)}dx=(\frac{x}{2})\sqrt{x(1-x)}+\frac{1}{8}sin^{-1}(2x-1)+c)\int_{0}^{1}$$
$$=\frac{1}{8}[sin^{-1}(1)-sin^{-1}(-1)]$$
$$=\frac{1}{8}[\frac{\pi}{2}-(\frac{-\pi}{2})]$$
$$=\frac{1}{8}(\pi)$$
$$\int_{0}^{1}\sqrt{x(1-x)}dx=\frac{\pi}{8}$$

$$\displaystyle \int_{\pi}^{\dfrac{3\pi}{4}}\dfrac{sin   2x}{cos^4  x  +  sin^4  x}dx=\displaystyle \int_{\pi}^{\dfrac{3\pi}{4}}\dfrac{d  sin^2  x}{(1-sin^2  x)^2+sin^4x}$$

$$=\displaystyle \int_{\pi}^{\dfrac{3\pi}{4}}\dfrac{d  sin^2  x}{(2  sin^2  x-2 sin^2  x +\dfrac{1}{2}+\dfrac{1}{2})} $$

$$sin^2  x=t$$$$\displaystyle \int_{0}^{\dfrac{1}{2}}\dfrac{dt}{(\sqrt{2}t-\dfrac{1}{\sqrt{2}})^2+ (\dfrac{1}{\sqrt{2}})^2}=\dfrac{1}{\sqrt{2}}\dfrac{1}{(\dfrac{1}{\sqrt{2}})}tan^{-1}\left [ \dfrac{\left (\sqrt{2}t- \dfrac{1}{\sqrt{2}} \right )}{\dfrac{1}{\sqrt{2}}} \right ]_{0}^{\dfrac{1}{2}}$$

$$=tan^{-1}(2t-1)_{0}^{{1}/{2}}$$

$$=tan^{-1}(0)-tan^{-1}(-1)$$

$$=0-\left ( \dfrac{-\pi}{4} \right )$$

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

$$\int_{a}^{0}\dfrac{x^5dx}{\sqrt{a^2-x^2}}x=a  sin \theta$$
$$\int_{o}^{a}\dfrac{a^5sin ^5\theta}{a   cos  \theta}a cos \theta d  \theta$$
$$\int_{0}^{\dfrac{\pi}{2}}a^5  sin^5\theta  d  \theta=a^5\int_{0}^{\dfrac{\pi}{2}}sin ^5\theta  d  \theta$$
$$=a^5\left ( \dfrac{4}{5}\right )\left ( \dfrac{2}{3}\right )$$
$$=\dfrac{8a^5}{15}$$

$$\displaystyle \int_{0}^{1} cos^{-1}\left ( \dfrac{1-x^2}{1+x^2} \right )$$

$$x=tan \theta$$

$$dx =sec^2 \theta   \cdot  d  \theta$$

$$\displaystyle \int_{0}^{\dfrac{\pi}{4}}cos^{-1}(cos   2\theta)cos^2\theta   d\theta$$

$$\displaystyle \int_{0}^{\dfrac{\pi}{4}}2  \theta   sec^2 \theta   d \theta =2\displaystyle \int_{0}^{\dfrac{\pi}{4}}\theta   sec^2 \theta   d  \theta$$

$$2 [ \theta  tan \theta  +  log  |cos \theta| ]_{0}^{\dfrac{\pi}{4}}$$

$$2 \left [ (\dfrac{\pi}{4}-0)+ log (\dfrac{1}{\sqrt{2}})\right ]$$

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

$$=\dfrac{\pi}{2}-log  2$$