Evaluate-0-pi-4-dx-cos-3-x-2-sin-2x-

Question Number 17444 by Tinkutara last updated on 06/Jul/17

Evaluate : \underset{0}{\int^{\frac{π}{4}}} \frac{d x}{\cos^{3} x \sqrt{2 \sin 2 x}}

Answered by Arnab Maiti last updated on 10/Jul/17

= \int_{0}^{\frac{π}{4}} \frac{dx}{2 {(sinx)}^{\frac{1}{2}} {(\cos x)}^{\frac{7}{2}}}

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

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

put tanx = z \Rightarrow \sec^{2} x dx = dz

= \frac{1}{2} \int_{0}^{1} \frac{(1 + z^{2}) dz}{z^{\frac{1}{2}}}

= \frac{1}{\sqrt{2}} \int_{0}^{1} (z^{- \frac{1}{2}} + z^{\frac{3}{2}}) dz

= \frac{1}{2} {[{2 z}^{\frac{1}{2}} + \frac{2}{5} z^{\frac{5}{2}}]}_{0}^{1}

= \frac{1}{2} (2 + \frac{2}{5}) = 1 + \frac{1}{5} = \frac{6}{5}

Commented by 786 last updated on 07/Jul/17

Your answer is wrong .

Commented by Tinkutara last updated on 10/Jul/17

Yes . Thanks Sir!

Commented by Arnab Maiti last updated on 10/Jul/17

Is it right now ?