MHB Integral using trig substitution

[tmt1]

I have

$$\int_{}^{} \frac{1}{\sqrt{1 - x^2}} \,dx$$

I can let $x = \sin\left({\theta}\right)$ then $dx = cos(\theta) d\theta$

then:

$$\int_{}^{} \frac{cos(\theta) d\theta}{\sqrt{1 - (\sin\left({\theta}\right))^2}}$$

Using the trig identity $1 - sin^2\theta = cos^2\theta$, I can simplify this to:

$$\int_{}^{} \frac{d\theta}{cos\theta}$$

so $ln|cos\theta| + C$ should be the answer.

Since $cosx = \sqrt{1 - sin^2 x}$, then it would be $ln|\sqrt{1 - sin^2 \theta}| + C$

I can substitute back in $x = \sin\left({\theta}\right)$, so

$$ln|\sqrt{1 - x^2}| + C$$

However, this is not the answer I have in the solutions.

 

[MarkFL]

Note that:

$$\sqrt{1-\sin^2(\theta)}\ne\cos^2(\theta)$$

Also, even if this were to be correct:

$$\int\frac{d\theta}{\cos(\theta)}$$

We would still have:

$$\int\frac{d\theta}{\cos(\theta)}\ne\ln\left|\cos(\theta)\right|+C$$

We can see this as:

$$\frac{d}{d\theta}\left(\ln\left|\cos(\theta)\right|+C\right)=-\tan(\theta)\ne\frac{1}{\cos(\theta}$$

Anyway, after your substitution, we have:

$$\int\,d\theta=\theta+C$$

Now, back-substitute for $\theta$...what do you get?