Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\displaystyle \int \frac{\tan(\ln(x^5))}{x}\, dx$. Please help me evaluate this! I'm having trouble figuring out what "u" should be using "u" substitution.

share|improve this question
1  
Write $\ln(x^5)$ as $5 \ln x$ and use $u = \ln x$. –  Umberto P. Dec 3 '13 at 21:26

1 Answer 1

Put $$ u =\ln(x^5) = 5 \ln x $$ So we put $$u = 5 \ln x\implies du = \dfrac 5x\,dx \implies \frac {du}{5} = \frac {dx}{x}$$

Then $$\int \frac{\tan(\ln(x^5))}{x}\, dx = \frac 15\int \tan u\,du\tag{1}$$

Integrating $\int tan u \,du$ is fairly straightforward, if we rewrite $\tan u = \dfrac{\sin u}{\cos u}$:

$$\frac 15\int \tan u\,du = \frac 15 \int \dfrac{\sin u}{\cos u}\,du\tag{2}$$ and then set $w = \cos u \implies dw = -\sin u \,du \iff -dw = \sin u \,du$. Substituting into $(2)$ gives us $$ \frac 15 \int \dfrac{\sin u}{\cos u}\,du = -\frac 15 \int \frac{dw}{w} = -\frac 15\ln|w| + C$$

Now it's just a matter of back substituting $w = \cos u$, and then $u = 5\ln x$ to get our final result as a function of $x$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.