# Math Help - Trig Integration by Parts Problem - # 4

1. ## Trig Integration by Parts Problem - # 4

This one doesn't seem like an integration by parts problem.

Integration by parts $\int udv = uv - \int vdu$

$\int 3\cos \sqrt{x} dx$

$\int 3\cos x^{1/2} dx$

What is the next step? What is u? What is dv?

2. ## Re: Trig Integration by Parts Problem - # 4

Write it out like this:

\displaystyle \begin{align*} \int{ 3\cos{ \left( \sqrt{x} \right) } \, \mathrm{d}x} &= 6 \int{ \sqrt{x} \cos{ \left( \sqrt{x} \right) } \, \frac{1}{2\sqrt{x}}\,\mathrm{d}x } \end{align*}

Then make the substitution \displaystyle \begin{align*} X = \sqrt{x} \implies \mathrm{d}X = \frac{1}{2\sqrt{x}}\,\mathrm{d}x \end{align*} and the integral becomes

\displaystyle \begin{align*} 6\int{ \sqrt{x}\cos{ \left( \sqrt{x} \right) } \, \frac{1}{2\sqrt{x}} \, \mathrm{d}x } &= 6 \int{ X \cos{(X)}\,\mathrm{d}X } \end{align*}

and now you can apply integration by parts.