Review question

# Can we carry out these three tricky integrals? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8778

## Solution

1. Find
1. $\displaystyle \int x \sqrt{2x-1} \:dx$,
Let $u = 2x - 1$, so that $du = 2 \: dx$. Then \begin{align*} \int x \sqrt{2x-1} \:dx = \int \frac{u+1}{2} \sqrt{u} \:\frac{du}{2} &= \frac{1}{4} \int u^{3/2} + u^{1/2} \:du \\ &= \frac{1}{4} \left( \frac{u^{5/2}}{\frac{5}{2}} + \frac{u^{3/2}}{\frac{3}{2}} + C \right) \\ &= \frac{1}{10} u^{5/2} + \frac{1}{6} u^{3/2} + C\\ &= \frac{1}{10} (2x-1)^{5/2} + \frac{1}{6} (2x-1)^{3/2} + C,\\ \end{align*}

where $C$ is a constant of integration.

1. $\displaystyle \int \frac{1}{\sqrt{2x - x^2}} \:dx$.
We can write $\begin{equation*} 2x - x^2 = -(x^2 - 2x) = -\left( (x-1)^2 - 1 \right) = 1 - (x-1)^2 \end{equation*}$ so that $\begin{equation*} \int \frac{1}{\sqrt{2x - x^2}} \:dx = \int \frac{1}{\sqrt{1 - (x-1)^2}} \:dx. \end{equation*}$

Take $\sin \theta = x - 1$, so that $\cos \theta \: d\theta = dx$ and \begin{align*} \int \frac{1}{\sqrt{1 - (x-1)^2}} \:dx = \int \frac{\cos \theta}{\sqrt{1 - \sin^2 \theta}} \:d\theta &= \int \frac{\cos \theta}{\cos \theta} \:d\theta \\ &= \theta + C \\ &= \sin^{-1} (x-1) + C, \end{align*}

where $C$ is a constant of integration.

1. Evaluate $\displaystyle \int_0^{\pi/4} \sec^4 x \:dx$.
We have that $\begin{equation*} \tan^2 x + 1 = \sec^2 x. \end{equation*}$ By the standard result (which can be proved using the quotient rule), we have that $\begin{equation*} \frac{d}{dx} \tan x = \sec^2 x. \end{equation*}$ and therefore $\begin{equation*} \int \sec^4 x \:dx = \int \sec^2 x \sec^2 x\:dx = \int \sec^2 x\tan^2 x + \sec^2 x \:dx. \end{equation*}$

The second term we can integrate by the standard result above. For the first term, we substitute $u=\tan x$ and get $\int \sec^2 x\tan^2 x\:dx = \int u^2\:du = \frac{\tan^3 x}{3}+C.$ Thus we have $\int_0^{\pi/4} \sec^4 x \:dx = \left[\frac{\tan^3 x}{3}+\tan x\right]_0^{\pi/4} = \frac{4}{3}.$