Evaluate ∫ x sec^2 x dx.

Question 87

HardOriginal question

Evaluate $\int x \sec^{2} x\,dx$ .

x\tan x + \log\left|\sec x\right| + C

Correct answerYour answer

x\tan x - \log\left|\sec x\right| + C

Correct answerYour answer

x \tan x + C

Correct answerYour answer

-\left(x\tan x - \log\left|\sec x\right|\right) + C

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

By parts, $\int u\,dv = uv - \int v\,du$ , with $u = x$ and $dv = \sec^{2} x\,dx$ , so $v = \frac{\sin x}{\cos x}$ .
$\frac{x \sin x}{\cos x} - \int \tan x\,dx = x\tan x - \log\left|\sec x\right| + C$ .

Worth noting

Adding the second integral instead of subtracting it is the slip one of the options is built on.

Ten questions from Indefinite Integrals, at your own pace.