# Question Video: Evaluate an Integral Using Inverse Trigonometric Functions Mathematics • Higher Education

Evaluate ∫_(0)^(1) 3/√(4 − 𝑥²) d𝑥.

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### Video Transcript

Evaluate the integral from zero to one of three divided by the square root of four minus 𝑥 squared with respect to 𝑥.

The question is asking us to evaluate a definite integral. And we’ve seen integrals in this form before. We know for a nonzero constant 𝑎, the integral of one divided by the square root of 𝑎 squared minus 𝑥 squared with respect to 𝑥 is equal to the inverse sin of 𝑥 divided by 𝑎 plus the constant of integration 𝐶. And we can see our integral is similar to this.

First, four can be written as two squared. Next, we want the numerator of our integral to be the constant one. So, we’ll need to take our constant of three outside of our integral. Doing both of these, we get three times the integral from zero to one of one divided by the square root of two squared minus 𝑥 squared with respect to 𝑥. And this integral is now in exactly the same form as our integral rule. Our value of 𝑎 is equal to two. So, we apply this rule with 𝑎 equal to two. Remember, we’re calculating a definite integral.

This gives us three times the inverse sin of 𝑥 over two evaluated at the limits of our integral, 𝑥 is equal to zero and 𝑥 is equal to one. The last thing we need to do is evaluate this at the limits of our integral. This gives us three multiplied by the inverse sin of one over two minus the inverse sin of zero over two. We can simplify this to give us three times the inverse sin of one minus the inverse sin of zero.

Finally, we can evaluate both of these expressions. The inverse sin of one is 𝜋 by six and the inverse sin of zero is zero. So, this gives us three times 𝜋 by six, which we can calculate to give us 𝜋 by two. And this is our final answer.

Therefore, we’ve shown the integral from zero to one of three divided by the square root of four minus 𝑥 squared with respect to 𝑥 is equal to 𝜋 by two.

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