Question

how to do this. find the antiderivative of ∫(sin^2 3x + cos 3x)^2 dx by using substitution ?

Answer 1

http://www.wolframalpha.com/input/pod.jsp?id=MSP187119ib7e8e4a715cg6000030a2112d0b8f4h7e&s=50&button=1

Answer 2

Square it out giving:
\[\int\limits \sin^4(3x)+\cos^2(3x)+2\sin^2(3x)\cos(3x)dx\]
Using some identities to get:
\[\int\limits \left( \frac{1}{2}(1-\cos(6x)) \right)^2dx+\int\limits \left( \frac{1}{2} (1+\cos(6x)) \right)dx+\frac{2}{3}\int\limits u^2 du\]
Using:
\[\sin^4(3x)=(\sin^2(3x))^2=(\frac{1}{2}(1-\cos(2(3x)))^2; \cos^2(3x)=\frac{1}{2}(1+\cos(2(3x))\]\[ u=\sin(3x); du=3\cos(3x)\]