# Need help

#### Moistwh

##### New member
How should i evaluate the following integral?

My teacher told me to use x = sec(t)/6 as an inverse trig substitution, but i end up with (secx)^3/216(tanx) and do not know how to proceed
Any hints?

#### MarkFL

##### Super Moderator
Staff member
Okay, using the suggested substitution:

$$\displaystyle x=\frac{\sec(t)}{6}\implies dx=\frac{1}{6}\sec(t)\tan(t)\,dt$$

We then get:

$$\displaystyle I=\frac{1}{6^4}\int_0^{\frac{\pi}{4}} \sec^4(t)\,dt$$

Can you proceed?

#### MarkFL

##### Super Moderator
Staff member
I would continue by using a Pythagorean identity as follows:

$$\displaystyle I=\frac{1}{6^4}\int_0^{\frac{\pi}{4}} \left(\tan^2(t)+1\right)\sec^2(t)\,dt$$

Let:

$$\displaystyle u=\tan(t)\implies du=\sec^2(t)\,dt$$

And we have:

$$\displaystyle I=\frac{1}{6^4}\int_0^{1} u^2+1\,du=\frac{1}{972}$$

#### lookagain

##### Senior Member
How should i evaluate the following integral? View attachment 11396

My teacher told me to use x = sec(t)/6 as an inverse trig substitution, but i end up with (secx)^3/216(tanx) and do not know how to proceed
Any hints?
I would advise you stay with MarkFL's posts

However, consider another way such as this:

Rewrite the integrand as:

$$\displaystyle x^2(36x^2 - 1)^{-1/2}(x \ dx)$$

Let $$\displaystyle \ u = 36x^2 - 1$$

$$\displaystyle du = 72x \ dx \ \implies$$

$$\displaystyle x \ dx = \dfrac{1}{72}du$$

$$\displaystyle u = 36x^2 - 1 \ \implies$$

$$\displaystyle u + 1 = 36x^2 \ \implies$$

$$\displaystyle \dfrac{u + 1}{36} = x^2$$

With $$\displaystyle \ u = 36x^2 - 1, \ \$$ when $$\displaystyle \ x = \dfrac{\sqrt{2}}{6}, \ \ u = 1$$.

With $$\displaystyle \ u = 36x^2 - 1, \ \$$ when $$\displaystyle \ x = \dfrac{1}{6}, \ \ u = 0$$.

The integrand is $$\displaystyle \ \ \bigg(\dfrac{u + 1}{36}\bigg)\bigg(u^{-1/2}\bigg)\bigg(\dfrac{1}{72}du\bigg)$$.

The integral is

$$\displaystyle \dfrac{1}{36(72)}\displaystyle\int_0^1(u + 1)u^{-1/2}du \ =$$

$$\displaystyle \dfrac{1}{2592}\displaystyle\int_0^1(u^{1/2} + u^{-1/2})du$$

.