In this question x x\,x and θ \theta\,θ are measured in radians.
Find ∫cosx dx\displaystyle\int \cos x\,dx∫cosxdx.
You may use the result ddx(tankx)=ksec2kx\dfrac{d}{dx}(\tan kx) = k\sec^2 kxdxd(tankx)=ksec2kx. Find ∫2sec24x dx\displaystyle\int 2\sec^2 4x\,dx∫2sec24xdx.
Hence, using the identity tan2θ≡sec2θ−1\tan^2\theta \equiv \sec^2\theta - 1tan2θ≡sec2θ−1, find ∫tan2θ dθ\displaystyle\int \tan^2\theta\,d\theta∫tan2θdθ.