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$$\int_0^{\frac{\pi}{4}}\frac{\sec x}{1+2\sin^2x}dx=?$$

Attempt:

$$=\int_0^{\frac{\pi}{4}}\frac{\sec x}{1+(1-\cos 2x)}dx$$ $$=\int_0^{\frac{\pi}{4}}\frac{\sec x}{2-\cos 2x}dx$$ $$=\sqrt{2}\int_0^{\frac{\pi}{4}}\frac{1}{\sqrt{1+\cos 2x}(2-\cos 2x)}dx$$ $$=\frac{\sqrt{2}}{3}\int_0^{\frac{\pi}{4}}\frac{1+\cos 2x+2-\cos 2x }{\sqrt{1+\cos 2x}(2-\cos 2x)}dx$$ $$=\frac{\sqrt{2}}{3}\int_0^{\frac{\pi}{4}}\frac{\sqrt{1+\cos 2x}}{(2-\cos 2x)}dx+\frac{\sqrt{2}}{3}\int_0^{\frac{\pi}{4}}\frac{1}{\sqrt{1+\cos 2x}}dx$$ The second integrand is easy to calculate. How can I solve the first integrand?

(I would also like to know if there are easier methods)

Aditya Dev
  • 4,774
  • if you use substitution $u=\sin x$ you end up with $1/(1-u^4)$ integrand, which is a lot more do-able. I'm not sure how you would compute the first integrand in your last line though – user160738 Dec 12 '15 at 04:17

1 Answers1

11

Hint:

$$\begin{align} I &=\int_{0}^{\frac{\pi}{4}}\frac{\sec{\left(x\right)}}{1+2\sin^{2}{\left(x\right)}}\,\mathrm{d}x\\ &=\int_{0}^{\frac{\pi}{4}}\frac{\cos{\left(x\right)}}{\cos^{2}{\left(x\right)}\left(1+2\sin^{2}{\left(x\right)}\right)}\,\mathrm{d}x\\ &=\int_{0}^{\frac{\pi}{4}}\frac{\cos{\left(x\right)}}{\left(1-\sin^{2}{\left(x\right)}\right)\left(1+2\sin^{2}{\left(x\right)}\right)}\,\mathrm{d}x\\ &=\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\mathrm{d}y}{\left(1-y^{2}\right)\left(1+2y^{2}\right)};~~~\small{\left[x=\arcsin{\left(y\right)}\right]}\\ \end{align}$$

Presumably you can take it from there.

David H
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