Doing Integration

Question

We have to integrate the following

$$\int \frac{a^2 \sin^2(x) + b^2\cos^2(x)}{a^4\sin^2(x)+b^4\cos^2(x)}dx$$

In this I divided both in Nr and Dr by $\cos^2 x$ . But stuck how to proceed .

Answer 1

\begin{align} \int \frac{a^2 \sin^2(x) + b^2\cos^2(x)}{a^4\sin^2(x)+b^4\cos^2(x)}dx&=\int\frac{a^2\tan^2(x)+b^2}{a^4\tan^2(x)+b^4}dx \end{align}

Letting $\frac{b}{a}u=\tan(x)$, then $\frac{b}{a}du=\sec^2(x)dx\implies dx=\frac{b}{a}\frac{du}{u^2+1}$. This gives us

\begin{align} \int \frac{a^2 \sin^2(x) + b^2\cos^2(x)}{a^4\sin^2(x)+b^4\cos^2(x)}dx&=\frac{b}{a}\int\frac{du}{u^2+1}\frac{b^2(u^2+1)}{b^2(a^2u^2+ b^2)}\\ &=\frac{b}{a}\int\frac{du}{a^2u^2+b^2}\\ &=\frac{b}{a^3}\int\frac{du}{u^2+\frac{b^2}{a^2}}\\ &=\frac{1}{a^2}\int\frac{\frac{b}{a}}{u^2+\frac{b^2}{a^2}}\\ &=\frac{1}{a^2}\arctan\left(\frac{au}{b}\right)+C\\ &=\frac{1}{a^2}\arctan\left(\frac{a^2}{b^2}\frac{b}{a}u\right)+C\\ &=\frac{1}{a^2}\arctan\left(\frac{a^2}{b^2}\tan(x)\right)+C \end{align}

Answer 2

Hint (quite big):

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We can write: $$I =\int \frac {(a^2-b^2)\sin ^2 x +b^2}{a^4 \sin ^4 x +b^4\cos ^4 x} dx =\int \sec^2 x \frac {a^2\tan ^2x +b^2}{a^4\tan ^4 x+b^4} dx$$ Now we have to substitute $u=\tan x $ and perform partial fraction decomposition to get $$I =\int \frac {1}{2 (a^2u^2+\sqrt {2}abu +b^2)} du +\int \frac {1}{2 (a^2u^2-\sqrt {2}abu +b^2)} du $$ Now we can integrate both of them by completing the square. Hope it helps.