Question Video: Integrating Trigonometric Functions Involving Reciprocal Trigonometric Functions Mathematics • Second Year of Secondary School

Video Transcript

Determine the indefinite integral of negative three tan squared of eight 𝑥 times csc squared eight 𝑥 evaluated with respect to 𝑥.

This does at first look quite tricky. However, if we recall some of our trigonometric identities, we can change this into something more manageable. We know that tan 𝑥 is equal to sin 𝑥 over cos 𝑥 and that csc 𝑥 is equal to one over six 𝑥. We can therefore rewrite our integrand, that’s the function we want to integrate, as negative three times sin squared of eight 𝑥 over cos squared of eight 𝑥 times one over sin squared of eight 𝑥. And then we see that we can cancel the sin squared of eight 𝑥. We can take the factor of negative three outside our integral to make the next step easier. And we have negative three times the integral of one over cos squared eight 𝑥 evaluated with respect to 𝑥.

Now we know that one over cos 𝑥 is equal to sec 𝑥. So our integral becomes negative three times the integral of sec squared of eight 𝑥 with respect to 𝑥. But of course the integral of sec squared 𝑎𝑥 with respect to 𝑥 is one over 𝑎 tan of 𝑎𝑥 plus constant 𝑐. And so in our case where the constant 𝑎 is equal to eight, the integral of sec squared eight 𝑥 with respect to 𝑥 is one over eight times the tan of eight 𝑥 plus the constant 𝑐. We distribute our parentheses and we see that our answer is negative three-eighths times tan of eight 𝑥 plus a new constant, since we multiplied our original one by negative three. So let’s call that uppercase 𝐶.