# $\int\frac{\sin\left(x\right)}{\cos\left(x\right)}\,\mathrm{d}x$ by substitution

I'm trying to solve the following integral: $$\int\frac{\sin\left(x\right)}{\cos\left(x\right)}\,\mathrm{d}x$$ Using the substitution method with the substitution $u = \sin\left(x\right)$.

The exercise has two parts: the first one is using the substitution $u = \cos\left(x\right)$. No problem. I'm having difficulties with the second part, which is using the substitution $u = \sin\left(x\right)$.

I spent a couple of hours with the exercise before asking here, and after some trials I got this: $$\int f\left(g\left(x\right)\right)g'\left(x\right)\,\mathrm{d}x = \int f\left(u\right)\,\mathrm{d}u$$ $$g\left(x\right) = \sin\left(x\right)$$ $$g'\left(x\right) = \cos\left(x\right)$$ $$f\left(x\right) = \frac{x}{\cos^2\left(\arcsin(x)\right)} = \frac{x}{1 - x^2}$$ $$\int f\left(u\right)\,\mathrm{d}u = -\frac{1}{2}\log|1 - u^2| + C = -\frac{1}{2}\log|1 - \sin^2\left(x\right)| + C$$ $$1 - \sin^2\left(x\right) = \cos^2\left(x\right)$$ $$\int\frac{\sin\left(x\right)}{\cos\left(x\right)}\,\mathrm{d}x = -\frac{1}{2}\log|\cos^2\left(x\right)| + C = -\log|\cos\left(x\right)| + C$$ But it feels too complicated, $f\left(x\right)$ was really hard for me to discover. What am I missing?

• what is wrong with noticing $\frac{d \cos x}{dx} = -\sin x$ so that $\sin x dx = - d \cos x$?
– jim
Jun 29, 2017 at 20:35

$\displaystyle \int\left(\frac{\sin\left(x\right)}{\cos\left(x\right)}\right)\,\mathrm{d}x$

As you noticed, it's easy to let $u = \cos (x)$, and that's how I would recommend it. But suppose we weren't being as slick as you were, but we still wanted to let $u = \sin (x)$. Then the problem comes from $du = \cos (x)$, which isn't there.

But $\dfrac{1}{\cos x} = \dfrac{\cos x}{\cos^2 x}$, and combining $u = \sin x$ with $\cos^2 x = 1 - \sin^2 x$, we get that $\dfrac{\cos x }{\cos^2 x} = \dfrac{du}{1 - u^2}$.

Thus $\displaystyle \int \dfrac{\sin x}{\cos x}dx = \int \dfrac{udu}{1-u^2}$, which is what you called $\int f(x)$.

• Your solution is definitely better than mine. Thanks! Jul 15, 2012 at 18:58

What you did is fine. One can view it as carrying out the standard strategy when we have a product of integer powers of sines and cosines, with at least one of the powers odd. In this case, both powers are odd.

Note that $$\frac{\sin x}{\cos x}=\frac{\sin x\cos x}{\cos^2 x}=\frac{\sin x\cos x}{1-\sin^2 x}.$$

Then the substitution $u=\sin x$ transforms our integral to $$\int \frac{u}{1-u^2}\,du.$$

Remark: If we had instead something like $\int\frac{\sin^4 x}{\cos x}\,dx$, then the same strategy of substituting for $\sin x$ would work. The substitution $u=\cos x$ would be less attractive.