Use reduction formulas to evaluate the integrals in Exercises ...

Question

Use reduction formulas to evaluate the integrals in Exercises 41–50.

∫ 2 sin^2(t) sec^4(t) dt

Accepted Answer

Welcome back everyone. Evaluate the integral of 6 sin squared t sequent to the power of 6 tdt. For this problem, let's begin by analyzing the function. We can also factor out the constant to write 6 integral sin squared t sequent to the power of 6 t dt. What we're going to do is rewrite this expression in terms of tangent and sequent. So essentially let's consider sin squared t and let's rewrite it as sin squared. Of t divided by cosine squared of t multiplied by cosine squared of t. In other words, we're dividing and multiplying by cosine squared of t. So this gives us tangent squared of t. And essentially cosine square of t can be rewritten in terms of sequent. This is basically 1 divided by sequin square of t. So now we are getting tangent squared of t divided by 2 squared of t. Now let's go ahead and multiply by sequence to the power of 6 of TDT, and this will essentially give us 6 integral of tangent. Squared of t multiplied by sequence to the power of 6 divided by sin squared, well, that'll be s to the power of 4 of cdc. Let's label this integral with i and continue. So what we're going to do is simply use trigonometric identities. We're going to split sequence to the power of 4. Let's go ahead and do that. So we get 6 integral. Tangent squared of t multiplied by square of t and multiplied by another sequin squared of t followed by dt. From here we are going to use the identity sequin squared of t equals 1 plus tangent squared of t. So we get 6 integral. Tangent squared of c multiplied by 1 plus Tangent squared see. 2 squared cdc and from here let's go ahead and apply substitution. Let's suppose that u is Tangent sea. And then the derivative of u with respect to t is going to be the derivative of tangent, which is sin squared t. So that if we multiply both sides by dt, we can show that sequin squared c dt is equal to the u. So what is the new integral? Well, I can be written as 6. Integral of u2 multiplied by 1 + u2 multiplied by d. Now we're going to distribute to get 6 integral. Of u2 plus u to the power of 4 d u and we can now integrate using the power rule. We get 6 in parentheses. U cubed divided by 3 plus u to the power of 5 divided by 5. Plus a constant of integration c. And now let's go ahead and distribute 6 divided by 5 gives us 6/5 use the power of 5, which is tangent to the power of 5. C plus 6 divided by 3 is 2, so we're going to add 2. u cubed is going to be tangent cubed c plus a constant of integration c, and that's our final answer for this problem. Thank you for watching.

Use reduction formulas to evaluate the integrals in Exercises 41–50.∫ 2 sin^2(t) sec^4(t) dt

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Use reduction formulas to evaluate the integrals in Exercises 41–50.
∫ 2 sin^2(t) sec^4(t) dt