Q.By using the properties of definite integrals, evaluate the integral
Using the property , the given integral simplifies to .
The trick here is symmetry. When you see an integral from to with a ratio of sines and cosines, the substitution often turns the denominator into a mirror image of itself. This lets you add the original and transformed integrals, giving a simple result.
Let’s work through it.
- Define the integral. Let
- Apply the symmetry substitution. Use the property . Here , so replace by :
Recall the co-function identities:
and .
So the integral becomes
- Add the two forms. Now we have two expressions for :
and
Add them:
The integrand simplifies to (provided the denominator is never zero on , which it isn’t — both terms are non-negative and only vanish at the endpoints, but the sum is positive in between).
- Evaluate the simple integral.
Hence , so
A common mistake is to forget that the substitution changes the limits but the property handles that automatically — you don’t need to recompute them. Also, be careful: the exponent is fine here because the functions are well-defined and positive on .
This trick works for any integral of the form where is any function for which the substitution works — the answer is always , as long as the denominator never vanishes.
The value of the integral is .
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