Question

π/4∫ sin²x/1+sinx cosx dx ,Evaluate it.0

Answer 1

Answer:

cos^4x+cos^3 is the answer

Answer 2

correct answer is $I = \frac{\pi}{6\sqrt{3}} - \frac{1}{2}\log \frac{3}{2}$

Explanation:

Let Let $I = \int_{0}^{\frac{\pi}{4}}\frac{\sin^{2}x}{1 + \sin x \cos x}dx$

$=\int_{0}^{\frac{\pi}{4}}\frac{2\sin ^{2}}{2 + 2\sin x \cos x}dx$

$= \int_{0}^{\frac{\pi}{4}}\frac{1-\cos 2x}{2 + \sin 2x}dx$

$= \int_{0}^{\frac{\pi}{4}}\frac{1}{2 + \sin 2x} - \int_{0}^{\frac{\pi}{4}}\frac{\cos 2x}{2 + \sin 2x}$

$I = I_{1} - I_{2}$………………………………(i)

$I_{1} = \int_{0}^{\frac{\pi}{4}}\frac{1}{2 + \sin 2x}$

$I_{1} = \int_{0}^{\frac{\pi}{4}}\frac{1}{2 + \sin 2x}dx$

$= \int_{0}^{\frac{\pi}{4}}\frac{\sec ^{2}x}{1 + \frac{2\tan x}{1+ \tan^{2}x}}dx$

$= \int_{0}^{\frac{\pi}{4}}\frac{\sec^{2}x}{2(1+ \tan^2 x + \tan x)}dx$

$=(1/2) \int_{0}^{\frac{\pi}{4}}\frac{\sec^2 x}{(\frac{1}{2} + tan x)^{2} + \frac{3}{4}}dx$

Let, $Let \frac{1}{2} + tan x = t \Rightarrow \sec^{2} x dx = dt$

At $x = 0 \Rightarrow t = \frac{1}{2}$ and $x = \frac{\pi}{4} \Rightarrow t = \frac{3}{2}$

$\frac{1}{2} \int_{\frac{1}{2}}^{\frac{3}{2}}\frac{1}{t^{2} + (\frac{\sqrt3}{2})^{2}}dt$

$= \frac{1}{2}\times\frac{2}{\sqrt3}\left [ tan^{-1}\frac{2x}{\sqrt3} \right ]_{\frac{1}{2}}^{\frac{3}{2}}$

$= \frac{1}{\sqrt3}\left [ \frac{\pi}{3} - \frac{\pi}{6} \right ]$

$I_{1} = \frac{\pi}{6\sqrt{3}}$

Now, $I_{2} = \int_{0}^{\frac{\pi }{4}}\frac{cos2x}{2+sin 2x} dx$

$2 + \sin 2x=t \Rightarrow 2\cos 2x dx=dt \Rightarrow \cos 2x=\frac{1}{2}dt$

At $x = 0 \Rightarrow t = 0$ and $x = \frac{\pi}{4} \Rightarrow t = 3$

$= \frac{1}{2}\int_{2}^{3}\frac{1}{t}dt$

$= \frac{1}{2} \left [ \log t \right ]_{2}^{3}$

$I_{2} = \frac{1}{2}\log \frac{3}{2}$

On putting the value of $I_{1}$ and $I_{2}$ in equation (i)

Hence,  $I = \frac{\pi}{6\sqrt{3}} - \frac{1}{2}\log \frac{3}{2}$.