Question Video: Evaluate a Definite Integral Involving the Exponential Function and Trigonometric Functions

Evaluate ∫_(πœ‹/3)^(πœ‹) 𝑒^(2 sin πœƒ) cos πœƒ dπœƒ.

04:10

Video Transcript

Evaluate the integral from πœ‹ over three to πœ‹ of 𝑒 to the power of two sin πœƒ multiplied by the cos of πœƒ with respect to πœƒ.

We’re asked to evaluate an integral which is not in a standard form which we know how to integrate. There are several different ways of tackling this integral. The easiest way is to notice that the derivative of the sin of πœƒ with respect to πœƒ is equal to the cos of πœƒ which we’re actually multiplying by our composite function inside of our integral. This should remind us of integration by substitution while we set 𝑒 equal to the function sin πœƒ. To apply integration by substitution, we need 𝑔 prime to be continuous on the closed interval of our integral. And we know that 𝑔 prime in this case is the cos of πœƒ, which is continuous on the whole real numbers. So this is true.

Next, we need our function 𝑓 to be continuous on the range of 𝑔. 𝑓 is the outer function inside of our integrand, which, in our case, is 𝑒 to the power of two πœƒ. So we need 𝑒 to the power two πœƒ to be continuous on the range of 𝑔. However, 𝑒 to the power of two πœƒ is a standard exponential function which is continuous on the whole real numbers. So, this is also true. Therefore, we’ve shown we’re allowed to integrate this by using the substitution 𝑒 is equal to the sin of πœƒ. We differentiate both sides of our substitution with respect to πœƒ to get d𝑒 by dπœƒ is equal to the cos of πœƒ. And although d𝑒 by dπœƒ is not a fraction, when we’re using integration by substitution, it does behave a little bit like a fraction. Using this, we can get the equivalent statement that d𝑒 is equal to the cos of πœƒ dπœƒ.

We’re almost ready to use the substitution. We just need to find the new limits of our integral. To find the upper limit of our integral, we substitute πœƒ is equal to πœ‹ into our substitution. So, our upper limit is equal to the sin of πœ‹ which is equal to zero. Similarly, to find a new lower limit of our integral we need to substitute πœƒ is equal to πœ‹ over three into our substitution. Substituting πœƒ is equal to πœ‹ over three into our substitution 𝑒 is equal to the sin of πœƒ gives us the sin of πœ‹ divided by three. And this is a standard trigonometric result that we should memorize. It’s equal to the square root of three divided by two.

We’re now ready to integrate this by using the substitution 𝑒 is equal to the sin of πœƒ. First, we found that our lower limit is equal to the square root of three over two and the upper limit is equal to zero. Next, we replace our inner function, the sin of πœƒ, with 𝑒. Finally, we show that the cos of πœƒ dπœƒ is equal to d𝑒. This means we now need to evaluate the integral from the square root of three over two to zero of 𝑒 to the power of two 𝑒 with respect to 𝑒. And this is now in a standard form which we can integrate.

For any constants π‘Ž and 𝑛 where 𝑛 is not equal to zero, the integral of π‘Ž multiplied by 𝑒 to the power of 𝑛π‘₯ with respect to π‘₯ is equal to π‘Ž times 𝑒 to the power of 𝑛π‘₯ divided by 𝑛 plus our constant of integration 𝐢. We divide by the coefficient of our exponent. So, we integrate 𝑒 to the power of two 𝑒 to get 𝑒 to the power two 𝑒 over two, where we don’t include our constant of integration since it will cancel when we evaluate at the limits of our definite integral. Evaluating this at the limits of our integral gives us 𝑒 to the power of two times zero over two minus 𝑒 to the power of two times the square root of three over two divided by two.

We have two times zero is equal to zero. And then any number raised to the zeroth power is equal to one. And in our second term, we can simplify the exponent by canceling the shared factor of two. This gives us one-half minus 𝑒 to the power of the square root of three divided by two. Finally, we take out the shared factor of one-half from both of our terms. This gives us one-half multiplied by one minus 𝑒 to the power of the square root of three. Therefore, we’ve shown that the integral from πœ‹ over three to πœ‹ of 𝑒 to the power of two times the sin of πœƒ all multiplied by the cos of πœƒ with respect to πœƒ is equal to one-half multiplied by one minus 𝑒 to the power of the square root of three.

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