Integrals Test - 16

$$\int_{1}^{2}\left ( \dfrac{1+x  \log  x}{x} \right ) e^x  dx$$

$$\int_{1}^{2}\left ( \dfrac{1}{x}  +  \log  x \right ) e^xdx$$           $$\displaystyle \int e^x [f(x)+f'(x) ] dx=e^x f(x)$$

$$= \int_{1}^{2} \left ( e^x  \log x+ c\right )  $$

$$=e^2  \log 2-e   \log  1$$

$$=e^2  \log  2-0$$

$$\int_{1}^{2}\left ( \dfrac{1+x  \log  x}{x} \right ) e^x  dx=e^2  \log  2$$

$$\displaystyle \int_{\frac{\sqrt{2}}{3}}^{\frac{\sqrt{3}}{3}} \dfrac{dx}{2\sqrt{1-\dfrac{ax^2}{2}}}=\dfrac{1}{2}\displaystyle \int_{\frac{\sqrt{2}}{3}}^{\frac{\sqrt{3}}{3}} \dfrac{dx}{\sqrt{1-(\dfrac{3x}{2})^2}}$$

$$=\left[\dfrac{2}{3}\times \dfrac{1}{2}  sin^{-1}\left(\dfrac{3x}{2}\right)\displaystyle \right]_{\frac{\sqrt{2}}{3}}^{\frac{\sqrt{3}}{3}}$$

$$=\dfrac{2}{3}\times \dfrac{1}{2} \left [ sin^{-1} \left ( \dfrac{3\sqrt{3}}{3(2)} \right ) -sin^{-1} \left ( \dfrac{3\sqrt{2}}{3\times 2} \right ) \right ]$$

$$=\dfrac{1}{3} \left [ sin^{-1} \left ( \dfrac{\sqrt{3}}{2} \right ) -sin \left ( \dfrac{1}{\sqrt{2}} \right ) \right ]$$

$$=\dfrac{1}{3} \left [ \dfrac{\pi}{3}-\dfrac{\pi}{4} \right ]$$

$$=\dfrac{1}{3} \left ( \dfrac{\pi}{12} \right )$$

$$=\dfrac{\pi}{36}$$

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

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

$$\displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{2  d  (tan \dfrac{x}{2})}{1-(tan\dfrac{x}{2}+1)^2}=2 \displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{d (tan \dfrac{x}{2})}{(\sqrt{2})^2-(tan \dfrac{x}{2}-1)^2}$$

$$=2\left [ \dfrac{1}{2\sqrt{2}} \right ]   log   \left | \dfrac{\sqrt{2}+tan\dfrac{x}{2}-1}{\sqrt{2}-(tan\dfrac{x}{2}-1)} \right | _{0}^{\dfrac{\pi}{2}}$$

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

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

$$=\sqrt{2}log (\sqrt{2}+1)$$

$$\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^5x\cdot sin2x dx=\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^5x\cdot 2 sinx   cosx   dx$$

$$=2\displaystyle \int_{0}^{\dfrac{\pi}{2}}  cos^6 x   sinx  dx=-2\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^6 x\cdot d(cos  x)$$

$$=-2\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^6  x   dcos  x=-2\left ( \dfrac{cos^7x}{7} \right )\displaystyle \vert_{0}^{\dfrac{\pi}{2}}$$

$$=-2\left [ \dfrac{cos^7  x}{7}\displaystyle \vert_{0}^{\frac{\pi}{2}} \right ]$$

$$=-2\left ( \dfrac{-1}{7} \right)=\dfrac{2}{7}$$

Thus $$\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^5  x\times sin 2x  dx=\dfrac{2}{7}$$

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

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

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

$$=(\sin^{-1}1+c)-(\sin^{-1}0+c)$$

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

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

Consider, $$I=\displaystyle \int_{0}^{\dfrac{\pi}{4}}\sqrt{\dfrac{1-sin 2  x}{1+sin 2  x}}\cdot dx$$

$$\displaystyle I= \int_{0}^{\dfrac{\pi}{4}}\sqrt{\dfrac{(sinx - cos x)^2}{(sin x + cos x)^2}}\cdot  dx$$

$$\displaystyle I=\int_{0}^{\dfrac{\pi}{4}}{\dfrac{(sinx - cos x)}{sin x + cos x}}\cdot dx$$

$$\displaystyle I=-\int_{0}^{\dfrac{\pi}{4}}{\dfrac{cos  x-  sin  x}{sin  x +  cos  x}}dx$$

$$I=-\left[log |sin  x  +  cos  x|\right]_0^{\frac{\pi}{4}}$$

$$I=-\left[log (sin  {\frac{\pi}{4}}  +  cos  {\frac{\pi}{4}}) -\log (\sin 0+\cos 0))\right]$$

$$I=-\left[log \left (\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt 2}\right)-\log 1\right]$$

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

So,  $$\displaystyle \int_{0}^{\dfrac{\pi}{4}}\sqrt{\dfrac{1- sin 2 x}{1+sin 2 x}}x=-log (\sqrt{2})$$

 $$\displaystyle \int_{0}^{1}\dfrac{(sin^{-1}x)^2}{\sqrt{1-x^2}} \cdot dx$$

$$\displaystyle \int \dfrac{(sin^{-1}x)^2}{\sqrt{1-x^2}} \cdot dx=\displaystyle \int (sin^{-1}x)d(sin^{-1}x)$$

$$=\dfrac{(sin^{-1}x)^3}{3}+c$$

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

$$=\left ( \dfrac{(sin^{-1}1)^3}{3}+c\right )-\left ( \dfrac{(sin^{-1}0)^3}{3}+c\right )$$

$$=\dfrac{1}{3}\left ( \dfrac{\pi}{2} \right )^3$$

$$=\dfrac{\pi^3}{24}$$

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

We know that $$\displaystyle \int \frac{1}{\sqrt{1-x^2}}dx=(\sin^{-1}x+c)$$

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

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

$$=\sin^{-1}1-\sin^{-1}\dfrac{1}{2}$$

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

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

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

$$\int_{0}^{1}\sqrt{1-x^2}\cdot dx$$
$$\int \sqrt{1-x^2}dx=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2} sin^{-1}\left ( \dfrac{x}{a} \right )+ c $$
$$\int_{0}^{1}\sqrt{1-x^2}\cdot dx= \left ( \dfrac{1}{2} \sqrt{1^2-x^2}+\dfrac{a^2}{2}sin^{-1} \left ( \dfrac{x}{a} \right )+c \right )\int_{0}^{1}$$
$$=\left ( \dfrac{1}{2} (0)+ \dfrac{1}{2} sin^{-1}(1)+c \right ) - (0+0+c)$$
$$=\dfrac{1}{2}\dfrac{\pi}{2}$$
$$=\dfrac{\pi}{4}$$
$$\int_{0}^{1}\sqrt{1-x^2}dx=\dfrac{\pi}{4}$$