How to solve Trigonometric Substitution questions?

Question

In this question , I could solve until that is seen on the photo. I don't know how to continue this question. What can I do it in order to continue? If you share a lot of solutions, I appreciate you :)

Answer 1

To solve the last integral you should notice that $$ d(\sec \theta) = \tan \theta \sec \theta d\theta $$ so your integral is $$ \int \tan^2\theta d(\sec \theta) = \int (\sec^2\theta - 1)d(\sec \theta) $$

edit:

For more colour to the answer. We can note that if we take the sub $$ u = \sec \theta $$ we obtain $$ du = \tan\theta\sec \theta d\theta $$ with the latter being embedded in the integral we are trying to solve. We also use the identity you used originally $$ \tan^2\theta + 1 = \sec^2\theta = u^2 $$ from there we have $$ \int (u^2 -1) du $$ which you can solve easily.

I know it seems a tad circular, but it is easier to process than trialing a function such as $$ x = 2\tan\left(\sec^{-1} u\right) $$ (unless you have seen it before).

Answer 2

Here's the usual procedure:

1. Notice if you have $x^2$ and $a^2$ with the same sign, or the opposite sign.
2. If they have the same sign, you'll typically use $x=\tan(\theta).$
3. If they have the opposite sign, make the positive term the hypotenuse, and the negative term one of the sides. You'll typically use $x=\sin(\theta)$ or $x=\cos(\theta).$
4. Draw your triangle, making $\theta$ one of the non-right angles. Label all sides, as well as $\theta.$
5. Use the triangle to write down the variable substitution, as well as, hopefully, a very simple relationship you can use to get the differential.
6. Substitute for the integrand, differential, and limits if there are any.
7. Perform the integration.
8. If there were limits, you should be done.
9. If there were no limits, you should use your relationships to transform back to the original variables.

So, for the integral $$\int\frac{x^3\,dx}{\sqrt{x^2+4}}, $$ we see that $x^2$ and $4$ have the same sign, which means we'll be using $\tan(\theta).$ The triangle will have $x$ as one side, $2$ as one side (we'll make it opposite $\theta$), and $\sqrt{x^2+4}$ as the hypotenuse. Then we can write down: \begin{align*} \frac{x}{2}&=\tan(\theta)\\ x&=2\tan(\theta)\\ dx&=2\sec^2(\theta)\,d\theta\\ \frac{2}{\sqrt{x^2+4}}&=\cos(\theta). \end{align*} Then we can write $$\int\frac{x^3\,dx}{\sqrt{x^2+4}}=\int 8\cos(\theta)\tan^3(\theta)\sec^2(\theta)\,d\theta=8\int\tan^3(\theta)\sec(\theta)\,d\theta. $$ Now this integral succumbs to the methods of integrating products of secants and tangents. The result is $$8\left[\frac{\sec^3(\theta)}{3}-\sec(\theta)\right]+C=\frac{(x^2+4)^{3/2}}{3}-4\sqrt{x^2+4}+C. $$ You can differentiate this to be certain that it is the correct answer.

Answer 3

$8\int tan^3 \theta$ $ sec \theta$ $ d\theta=8\int \frac{sin^3 \theta}{cos^3 \theta} \frac{1}{cos\theta}$ $ d\theta=8\int \frac{sin^3 \theta}{cos^4 \theta}d\theta$

$=8\int sin^3\theta$ $ cos^{-4}\theta$ $ d\theta$

$=8\int sin\theta$ $ sin^2\theta$ $ cos^{-4}\theta$ $ d\theta$

$=8\int sin\theta$ $ (1-cos^2\theta)$ $ cos^{-4}\theta$ $ d\theta$

$=8\int sin\theta$ $ cos^{-4}\theta - sin\theta$ $ cos^{-2}\theta$ $ d\theta$

$=8(\int sin\theta$ $ cos^{-4}\theta$ $ d\theta - \int sin\theta$ $ cos^{-2}\theta$ $ d\theta)$

and from here you can use u-substitution on both integrals to solve... let $u=cos\theta$