Integrate (cos(x)+(sqrt(1+x^2))*(sin(x)^3)*(cos(x)^3)) from -pi/4 to pi/4
[I know the solution (from wolfram alpha); I'm interested in the solution method. Gonna have to do this type of integral without wolfram]
The first thing I would do is separate the two parts: $(cos(x)+ \sqrt{1+ x^2})(sin^3(x)cos^3(x)= (sin^3(x))(cos^4(x)+ \sqrt{1+ x^2}(sin^3(x))(cos^3(x))$.
The first part $\int_{-\pii/4}^{\pi/4} sin^3(x)cos^4(x) dx$ is easy- since sin(x) is to an odd power, factor out one to use with the dx, then change the remaining to cos(x):
$\int_{-\pi/4}^{\pi/4} sin^2(x)cos^4(x) (sin(x)dx)= \int_{-\pi/4}^{\pi/4} (1- cos^2(x))(cos^4(x) (sin(x) dx)$
Now, let $u= cos(x)$ so that $du= -sin(x)dx$. When $x= \pi/4$, $u= \sqrt{2}/2$ and, since cosine is an even function, when $x= -\pi/2$, $u= \sqrt{2}/2$. The integral becomes $\int_{\sqrt{2}/2}^{\sqrt{2}/2} u^4- u^6 du$ and, since the limits of integration are the same, the integral is 0.
(More simply, since cos(x) is an even function and sin(x) is an odd function, sin(x) to an odd power is an odd function so the integrand is odd and the integral from any -a to a is 0.)
And, while there is no simple "anti-derivative for $\sqrt{1+ x^2} sin^3(x)cos^3(x)$. We can apply the same argument- $\sqrt{1+ x^2}$ is an even function, $cos^3(x)$ is an even function, and $sin^3(x)$ is an odd function so this integrand is an odd function. Its integral, from any -a to a is 0.