Video: Evaluating a Double Integral

Evaluate the double integral ∫_(0) ^(𝜋/2) ∫_(0) ^(𝑦) cos 𝑥 sin 𝑦 d𝑥 d𝑦.

05:07

Video Transcript

Evaluate the double integral, the integral between 𝜋 by two and zero of the integral between 𝑦 and zero of cos 𝑥 sin 𝑦 d𝑥 d𝑦.

Double integrals are a way to integrate over a two-dimensional area. Among other things, they let us work out the volume under a surface. As the name suggests, it involves integrating twice. In this case, our first step is to integrate cos 𝑥 sin 𝑦 with respect to 𝑥 between the limit zero and 𝑦. We’ll then integrate this answer with respect to 𝑦 between the limit zero and 𝜋 by two.

As stated, our first step is to integrate cos 𝑥 sin 𝑦 with respect to 𝑥. As sin 𝑦 contains no 𝑥 term, this can be treated as a constant. We, therefore, need to integrate cos 𝑥 with respect to 𝑥 and multiply it by the constant sin 𝑦. The integral of cos 𝑥 d𝑥 is a standard integral we need to remember. It is equal to the sin of 𝑥. This means that the integral of cos 𝑥 sin 𝑦 d𝑥 is sin 𝑥 sin 𝑦. As always with definite integrals, we now need to substitute in the upper limit and the lower limit and subtract our answers.

Substituting in the upper limit gives us sin 𝑦 multiplied by sin 𝑦. And substituting in the lower limit gives us sin of zero multiplied by sin of 𝑦. The sin of zero is equal to zero. Therefore, the second term equals zero. sin 𝑦 multiplied by sin 𝑦 can be rewritten as sin squared 𝑦. We have now completed the first part of the integration. The integral between 𝑦 and zero of cos 𝑥 sin 𝑦 d𝑥 is sin squared 𝑦.

We now need to integrate this answer with respect to 𝑦 between zero and 𝜋 by two. One of our double angle formulae states that cos of two 𝑥 is equal to one minus two sin squared 𝑥. This can be rearranged so that two sin squared 𝑥 is equal to one minus cos two 𝑥. By dividing both sides of this equation by two, sin squared 𝑥 is equal to one-half minus cos two 𝑥 over two. This means that we can rewrite the integral of sin squared 𝑦 as the integral of a half minus cos two 𝑦 over two. We do this as there is no standard integral for sin squared 𝑦.

Integrating a half with respect to 𝑦 gives us a half 𝑦 and integrating negative cos two 𝑦 over two gives us negative sin of two 𝑦 over four. This is because the integral of cos of 𝑛𝑥 is equal to sin of 𝑛𝑥 over 𝑛. We now need to substitute in the limits 𝜋 over two and zero and subtract our two answers. Substituting in 𝜋 over two gives us a half multiplied by 𝜋 over two minus sin of two multiplied by 𝜋 over two divided by four. One-half multiplied by 𝜋 over two is 𝜋 over four. sin of two multiplied by 𝜋 over two is the same as sin of 𝜋. The sin of 𝜋 is equal to zero. Therefore, the sin of 𝜋 divided by four is also equal to zero.

Substituting in our upper limit of 𝜋 over two gives us 𝜋 over four minus zero. This is equal to 𝜋 over four. Substituting in our lower limit of zero gives us a half multiplied by zero minus sin of two multiplied by zero divided by four. One-half multiplied by zero equals zero. Two multiplied by zero equals zero and the sin of zero also equals zero. This means that our lower limit of zero just gives us an answer of zero. 𝜋 by four minus zero equals 𝜋 by four.

We can, therefore, say that the final answer for the double integral is 𝜋 over four.

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