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Grade 12Integral Calculus

integrate the following:-
how to integrate the following function
limits are from 0 to pi/4
∫dx÷(cos^3x (2sin2x)1/2)

Profile image of Abhishek Sharma
10 Years agoGrade 12
Answers icon

1 Answer

Profile image of BALAJI ANDALAMALA
10 Years ago
\int_{0}^{\frac{\pi}{4}}\frac{dx}{cos^3x\sqrt{2sin2x}}=\int_{0}^{\frac{\pi}{4}}\frac{sec^2x.secx }{\sqrt{\frac{4tanx}{1+tan^2x}}}dx
=\int_{0}^{\frac{\pi}{4}}\frac{sec^2x.sec^2x }{\sqrt{{4tanx}}}dx
=\int_{0}^{\frac{\pi}{4}}\frac{sec^2x(1+tan^2x) }{2\sqrt{{tanx}}}dx
=\frac{1}{2}\int_{0}^{\frac{\pi}{4}}sec^2x(tan^{\frac{-1}{2}}x+tan^{\frac{3}{2}}x) dx

Put tanx = t , we get sec^2x dx = dt

\frac{1}{2}\int_{0}^{1}(t^{\frac{-1}{2}}+t^{\frac{3}{2}}) dt =\left [ t^{\frac{1}{2}}+\frac{t^{\frac{5}{2}}}{5} \right ]_{0}^{1}
=\frac{6}{5}

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