# Integrate 0 to π/2 (x sin x cos x )/(a²cos²x + b²sin²x)²

Abhishek Singh
105 Points
2 months ago
I think you can see the function is product of an algebraic and trig function indicating By-Parts.(Morover trig function is integrable)

Let us see the integral of trig part i.e $\frac{\sin x \cos x}{(a^2\cos ^{2}x + b^2\sin^2 x )^2}$ . We do this by subsituting denominator by t. i.e
$t = a^2\cos ^{2}x + b^2\sin^2 x \Rightarrow dt = 2(b^2-a^2)\sin x\cos x\\ \int \frac{sin x \cdot cos x dx }{(a^2\cos ^{2}x + b^2\sin^2 x )^2} = \frac{1}{2(b^2-a^2))}\int \frac{dt}{t^2} = \frac{-1}{2(b^2-a^2)t} = \frac{-1}{2(b^2-a^2)(a^2cos^2x+b^2sin^2x))}$

Now using By parts we get
\begin{align*} \int x\frac{sin x \cdot cos x\cdot dx }{(a^2\cos ^{2}x + b^2\sin^2 x )^2} &= x \left ( \frac{-1}{2\left ( b^2-a^2 \right)\left (a^2\cos ^{2}x + b^2\sin^2 x \right )} \right )-\int \left ( \frac{-1}{2\left ( b^2-a^2 \right)\left (a^2\cos ^{2}x + b^2\sin^2 x \right )} \right ) \\ &= x\frac{-1}{2\left ( b^2-a^2 \right)\left (a^2\cos ^{2}x + b^2\sin^2 x \right )} + \frac{1}{2(b^2-a^2))} \cdot \frac{tan^{-1}(\frac{b}{a}tanx)}{ab} \end{align*}Now we shall solve the second integral  using substitution t = tan
\begin{align*} \int \frac{dx}{a^2\cos ^{2}x + b^2\sin^2 x } &= \int \frac{sec^2x\cdot dx}{a^2 + b^2tan^2x}\\ &= \int \frac{dt}{a^2+b^2t^2} \\ &= \frac{1}{ab}tan^{-1}(bt/a) \\ &= \frac{1}{ab}tan^{-1}(\frac{btanx}{a})) \end{align*}
Now using this we get \begin{align*} \int x\frac{sin x \cdot cos x\cdot dx }{(a^2\cos ^{2}x + b^2\sin^2 x )^2} &= x \left ( \frac{-1}{2\left ( b^2-a^2 \right)\left (a^2\cos ^{2}x + b^2\sin^2 x \right )} \right )-\int \left ( \frac{-1}{2\left ( b^2-a^2 \right)\left (a^2\cos ^{2}x + b^2\sin^2 x \right )} \right ) \\ &= x\frac{-1}{2\left ( b^2-a^2 \right)\left (a^2\cos ^{2}x + b^2\sin^2 x \right )} + \frac{1}{2(b^2-a^2))} \cdot \frac{tan^{-1}(\frac{b}{a}tanx)}{ab} \end{align*}Now Applying limits we get
\begin{align*} \int_{0}^{\pi/2} x\frac{sin x \cdot cos x\cdot dx }{(a^2\cos ^{2}x + b^2\sin^2 x )^2} &= \frac{-\pi/2}{2\left ( b^2-a^2 \right)\left (b^2 \right )} + \frac{1}{2(b^2-a^2))} \cdot \frac{\pi/2-0}{ab} \\ &= \frac{\pi/2}{2(b^2-a^2)b}\cdot (\frac{1}{a}-\frac{1}{b}) \\ & =\frac{\pi/2}{2(b+a)ab^2} \end{align*}