# A question about Indefinite trigonometric integration

I am practising a unit on Integration. I am going through some past year papers, and there are some types of questions that I could not solve. So if anyone could help me in this, I'd really appreciate this.

Evaluate the following integrals: $$i)\quad\int\left(\frac{2}{\sqrt{x}}+2e^{-4x}+\frac{1}{3(1-x)}\right)\:\mathrm{d}x$$

In this, I managed to get to a point where the answer is: $$4\sqrt{x}-\frac{e^{-4x}}{2}+\frac{\ln|x+1|}{3}+C$$

Is this the final answer or there's more I can do here. I'm specially confused about the $|x+1|$ part.

Next is a trigonometric substituition question. I've tried basic ones of this type, but this one is very difficult and complicated for me. If someone could point me into a direction then maybe I can try solving.

Show that: $$\int\frac{x^2}{\sqrt{36-x^2}}\:\mathrm{d}x=18\sin^{-1}\left(\frac{x}{6}\right)-\frac{1}{2}x\sqrt{36-x^2}+C$$ with an appropriate trigonometric substitution.

• In $i)$, the last term is $-\frac{1}{3}\ln|1-x|$: note that your result is correct up to an arbitrary constant $C$. In the second integral why don't you try with $x=6\sin t$? Aug 19, 2013 at 17:18
• Added $\mathrm{\LaTeX{}}$ formatting, please refer to the starter guide on how to correctly typeset math on this site. In addition, don't consider the deletion of "Thanks in advance" an act of un-education, it's just a form to be coincise on the Question. Aug 19, 2013 at 17:31

• i)

The result has to be slightly modified; the last function is $-\frac{1}{3}\ln|1-x|$; in fact its derivative w.r.t. $x$ is equal to

$$-\frac{1}{3}\frac{-1}{1-x}=\frac{1}{3(1-x)},$$

if $1-x>0$ and

$$-\frac{1}{3}\frac{1}{x-1}=\frac{1}{3(1-x)},$$

if $1-x<0$.

Moreover, the result of the integral is given by the sum of the proposed functions (with the replacement of the third with the one above) plus an arbitrary constant $C$.

• ii)

With the substitution $x=6\sin t$, $dx=6\cos t dt$, the proposed integral is equal to

$$\int \frac{36\sin^2 t}{6\sqrt{1-\sin^2 t}}6\cos t dt=\int \frac{36\sin^2 t}{|\cos t|}\cos t dt= \int36\sin^2 t dt,$$

as $\cos t> 0$, if $t\in(-\frac{\pi}{2},\frac{\pi}{2})$, which is the taken image of $\operatorname{arcsin}(\cdot)$.

All you need is

$$36\int\sin^2 t dt=36\left(\frac{1}{2}t-\frac{1}{4}\sin(2t)\right)+C,$$

with $\sin(2t)=2\sin t\cos t$.

• you are welcome! Aug 20, 2013 at 8:58

Here we can put: $$x=6\sin\theta,\:\text{then we can get}\;\frac{\mathrm{d}x}{\mathrm{d}\theta}=6\cos\theta\:\mathrm{d}\theta$$ so that your integral becomes: $$\int\frac{36\sin^2\theta(6\cos\theta)}{6\cos\theta}\:\mathrm{d}\theta=\int 36\sin^2\theta\:\mathrm{d}\theta$$ I think, from this you should be able to proceed further.

EDIT: For your first question, yes that is the final answer except the fact that you made a small mistake. Always to check your answer, differentiate it and see if it's the same as your question.

ANS: $$4\sqrt{x}-\frac{e^{-4x}}{2}+\frac{\ln|1-x|}{3}+C$$

• appreciate the edit ,but the modulus expansion is a bit relevant to the problem... Aug 19, 2013 at 17:58
• This helped me solve the problems. Thanks. Aug 20, 2013 at 8:43
• Ok i have one final problem in this. I have already solved the whole question and I get the answer as 18(\Theta - (sin2\Theta)/2) and with that I can prove the first part tht is 18 sin^-1(x/6) but cannot get the second part of (1/2)x\sqrt{36-x^2}. Can some one please help me on how to get the secnd part? Aug 20, 2013 at 12:39
• we write the $sin2\theta\$ as = 2 * sin \theta\ * cos\theta* . then we assumed x/6 = sin \theta\ . proceed to find cos\theta\ and you are done!! Aug 20, 2013 at 13:43
• got it. Thanks :D Aug 20, 2013 at 14:31