Hello guys I'm totally lost in this indefinite integral, i'm just looking for advices/tips
$$\int \frac {\cos^5x}{ 16(\cos^4x+\sin^4x)}dx$$
Should I begin with universal substitution? or there is any other way to reduce the difficulty of this integral? As always thanks in advance!
On
Notice, $$\int \frac{\cos^5x}{16(\cos^4x+\sin^4x)}\ dx=\frac{1}{16}\int \frac{(1-\sin^2x)^2}{((1-\sin^2x)^2+\sin^4x)}\ (\cos x dx)$$ let $\sin x=t\implies \cos x\ dx=dt$, $$=\frac{1}{16}\int\frac{(1-t^2)^2}{((1-t^2)^2+t^4)}\ dt$$ $$=\frac{1}{16}\int\frac{t^4-2t^2+1}{2t^4-2t^2+1}\ dt$$
$$=\frac{1}{16}\int\left(\frac{1}{2}+\frac{\frac{1}{2}-t^2}{2t^4-2t^2+1}\right)\ dt$$ $$=\frac{1}{16}\left(\int \frac12 \ dt+\int\frac{\frac{1}{2t^2}-1}{2t^2-2+\frac{1}{t^2}}\ dt\right)$$
$$=\frac{1}{16}\left(\frac t2-\frac{1}{4}\int\frac{(\sqrt 2+1)\left(\sqrt 2-\frac{1}{t^2}\right)+(\sqrt 2-1)\left(\sqrt 2+\frac{1}{t^2}\right)}{2t^2+\frac{1}{t^2}-2}\ dt\right)$$
$$=\frac{t}{32}-\frac{1}{64}\int \frac{(\sqrt 2+1)\left(\sqrt 2-\frac{1}{t^2}\right)}{2t^2+\frac{1}{t^2}-2}\ dt-\frac{1}{64}\int \frac{(\sqrt 2-1)\left(\sqrt 2+\frac{1}{t^2}\right)}{2t^2+\frac{1}{t^2}-2}\ dt$$
$$=\frac{t}{32}-\frac{\sqrt 2+1}{64}\int \frac{\left(\sqrt 2-\frac{1}{t^2}\right)dt}{\left(\sqrt 2t+\frac{1}{t}\right)^2-2(\sqrt 2+1)}-\frac{\sqrt 2-1}{64}\int \frac{\left(\sqrt 2+\frac{1}{t^2}\right)dt}{\left(\sqrt 2t-\frac{1}{t}\right)^2+2(\sqrt 2-1)}$$
$$=\frac{t}{32}-\frac{\sqrt 2+1}{64}\int \frac{d\left(\sqrt 2t+\frac{1}{t}\right)}{\left(\sqrt 2t+\frac{1}{t}\right)^2-\left(\sqrt{2(\sqrt 2+1)}\right)^2}-\frac{\sqrt 2-1}{64}\int \frac{d\left(\sqrt 2t-\frac{1}{t}\right)}{\left(\sqrt 2t-\frac{1}{t}\right)^2+\left(\sqrt{2(\sqrt 2-1)}\right)^2}$$
$$=\frac{t}{32}-\frac{\sqrt 2+1}{64}\cdot \frac{1}{2\sqrt{2(\sqrt 2+1)}}\ln\left|\frac{\sqrt 2t+\frac{1}{t}-\sqrt{2(\sqrt 2+1)}}{\sqrt 2t+\frac{1}{t}+\sqrt{2(\sqrt 2+1)}}\right|-\frac{\sqrt 2-1}{64}\cdot \frac{1}{\sqrt{2(\sqrt 2-1)}}\tan^{-1}\left(\frac{\sqrt 2t-\frac{1}{t}}{\sqrt{2(\sqrt 2-1)}}\right)+C $$
$$=\color{red}{\frac{t}{32}-\frac{\sqrt{\sqrt 2+1}}{128\sqrt 2}\ln\left|\frac{\sqrt 2t+\frac{1}{t}-\sqrt{2(\sqrt 2+1)}}{\sqrt 2t+\frac{1}{t}+\sqrt{2(\sqrt 2+1)}}\right|-\frac{\sqrt{\sqrt 2-1}}{64\sqrt 2}\tan^{-1}\left(\frac{\sqrt 2t-\frac{1}{t}}{\sqrt{2(\sqrt 2-1)}}\right)+C }$$
On
Notice that all the trig functions in your integral have even powers, except for the cosine in the numerator. This tells us we can use the "extra" power of $\cos x$ with the $dx$ to get a good substitution. $\cos x\,dx$ is the differential of $\sin x$, so let's change all the other trig functions to sines.
$$\begin{align} \int \frac {\cos^5x}{ 16(\cos^4x+\sin^4x)}\,dx &= \int \frac {(\cos^2 x)^2}{16[(\cos^2 x)^2+\sin^4x]}\cos x\,dx \\ &= \int \frac {(1-\sin^2 x)^2}{16[(1-\sin^2 x)^2+\sin^4x]}\,d(\sin x) \\ &= \int \frac {(1-u^2)^2}{16[(1-u^2)^2+u^4]}\,du \end{align}$$
with $u=\sin x$. You now have a rational function in $u$ to integrate, which can be done by the usual means.
HINT: this integral has an elemetary antiderivative and you can use the tan half angle substitution