$\int_{0}^{\frac{\pi}{2}} \frac{1}{4\cos ^{2}x + 9\sin^{2}x} dx$
I was able to reduce it to $\int_{0}^{\frac{\pi}{2}} \frac{2}{17 - \cos2x} dx$ How do I proceed further ?
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Using the formula at this MSE link we obtain $$\frac{1}{4}\frac{2\pi}{6}= \frac{\pi}{12}.$$ – Marko Riedel Mar 26 '16 at 22:37
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Possible duplicate of Calculate the integral $\int_{0}^{2\pi}\frac{1}{a^{2}\cos^2t+b^{2}\sin^{2}t}dt$, by deformation theorem. – Guy Fsone Jan 20 '18 at 10:56
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Hint: $$\int_{0}^{\pi/2}\frac{1}{4\cos^2 x + 9 \sin^2 x}dx = \int_{0}^{\pi/2}\frac{\sec^2 x}{4 + 9 \tan^2 x}dx$$ Put $\tan x = t$ and solve.
TenaliRaman
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Beside the solutions you already have been given, I suppose that you made some mistake in your simplification since $$4\cos^2(x) + 9 \sin^2(x)=4\cos^2(x) + 4 \sin^2(x)+ 5 \sin^2(x)=4+5 \sin^2(x)$$ Now, using $$\sin^2(x)=\frac{1-\cos(2x)} 2$$ you should have ended with $$4\cos^2(x) + 9 \sin^2(x)=\frac{13-5\cos(2x)} 2$$ So $$I=\int\frac{dx}{4 \cos ^{2}(x) + 9 \sin^{2}(x)}=\int\frac{2dx}{13-5\cos(2x)}=\int\frac{dy}{13-5\cos(y)}$$ Now using the tangent half-angle substitution $t=\tan(\frac y2)$ $$I=\int\frac{dt}{4+9 t^2}=\frac 14\int\frac{dt}{1+\frac94t^2}=\frac 14 \times \frac 23\int\frac{du}{1+u^2}=\frac 16 \tan^{1}(u)$$ Now, concerning the bounds $$0\leq x \leq \frac \pi 2\implies 0\leq y \leq \pi \implies 0\leq t\leq\infty$$
Claude Leibovici
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HINT:
Divide the numerator & the denominator
by $4\cos^2x$ and set $\dfrac32\tan x=u$
or by $9\sin^2x$ and set $\dfrac23\cot x=v$
lab bhattacharjee
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