Recently, I've been into solving integration problems in my text book. The most tricky ones I find are those involving substitution method to solve them. How much ever I practice I cannot solve much of them; they seem to be very tricky.

I don't just get the way to work them out, I cannot figure out what term (usually trigonometric) to replace with what term in the problem. If I want to simplify an equation, I need to convert a trigonometric equation in the problem to a different equation. For example:


Are there tips to solve these type of problems ?

  • $\begingroup$ Defining $\sqrt x=y$ would help $\endgroup$ Feb 22, 2015 at 10:32

2 Answers 2


The equation you showed doesn't involve integration. Anyway, whenever you find some complicated expression (rational expression, to be precise) of trigonometric functions, Weierstrass substitution might be helpful.

This specific equation, however, is easy enough to be solved directly. Note that $$\sin\sqrt x + \cos\sqrt x = \sqrt 2 \sin\left(\sqrt x + \frac\pi 4\right)$$ so your equation becomes $$ \sin\left(\sqrt x + \frac \pi 4\right) = \frac 1{\sqrt 2}. $$ I guess you can proceed from here.


Consider chain rule but backwards.$$f[g(x)]dx=g'(x)f'[g(x)]$$Therefore:$$\int g'(x)f'[g(x)]dx=f[g(x)]$$What you want to do is to manipulate the integral with substitution, following that rule.

For example:$$\int f'(x)dx=f(x)$$You want to use the substitution $x=g(u)$.

To do that, you have to differentiate:$$xdx=g(u)du$$$$1=g'(u)$$$$\int(1)f'(x)dx=f(x)$$$$\int g'(x)f'[g(x)]dx=f[g(x)]$$This was always confusing for me at first, but looking at it, it made more sense for me. Also note the $(1)$ where we put $g'(x)$ because that is what $xdx=g(u)du$ gave us to substitute in.

Perhaps you wanted to use the following substitution:$$\sqrt{x^2-4}$$First, draw a triangle. Remember $a^2+b^2=c^2$. Try to make it look like the thing you want to substitute.$$a^2+b^2=c^2$$$$a^2=c^2-b^2$$$$a=\sqrt{c^2-b^2}$$$$a=g(x)=\sqrt{x^2-2^2}$$Now choose an angle to use.

Use trigonometric identities (mainly $\sin=\frac{opp}{hyp},\cos=\frac{adj}{hyp},\tan=\frac{opp}{adj}$)

Substitute them in (don't forget to differentiate and apply backwards chain rule) and integrate from there. If it didn't work, try again with another trigonometric functions, manipulate with some trigonometric identities, and you should eventually get something integratable.

Do mind me if I've made a few mistakes, I haven't actually been taught this at school yet, so I'm a bit rusty. (I self-teach)


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