Q. 1TF Trigonometric integrals: The int... [FREE SOLUTION]

Chapter 5: Q. 1TF (page 420)

Trigonometric integrals: The integrals that follow can be solved by using algebra to write the integrands in the form $f' (u (x)) u' (x)$ so that $u -$ substitution will apply.
Solve $鈭� \sin^{5} x d x$ by using the Pythagorean identity $\sin^{2} x + \cos^{2} x = 1$ to rewrite the integrand as role="math" localid="1649173494208" ${(1 鈭� \cos^{2} x)}^{2} \sin x$ and then applying substitution with $u = \cos x$ .

Short Answer

Expert verified

The value of integral is $鈭� \sin^{5} xdx = - \cos x + \frac{2}{3} \cos^{3} x - \frac{1}{5} \cos^{5} x + C$ .

Step by step solution

Step 1. Given Information

Solve $鈭� \sin^{5} x d x$ by using the Pythagorean identity $\sin^{2} x + \cos^{2} x = 1$ to rewrite the integrand as ${(1 鈭� \cos^{2} x)}^{2} \sin x$ and then applying substitution with $u = \cos x$ .

Step 2. The given integral is ∫sin5xdx.

We can write as

$鈭� \sin^{5} x d x = 鈭� {(1 鈭� \cos^{2} x)}^{2} \sin x d x$

Let

$u = \cos x \frac{d u}{d x} = - \sin x d u = - \sin x d x - d u = \sin x d x$

Step 3. Now the integral is

$鈭� \sin^{5} x d x = - 鈭� {(1 鈭� u^{2})}^{2} du 鈭� \sin^{5} xdx = - 鈭� {{{(1)}^{2} 鈭� 2 脳 1 脳 u^{2} + (u^{2})}^{2}} du 鈭� \sin^{5} xdx = - 鈭� (1 鈭� 2 u^{2} + u^{4}) du 鈭� \sin^{5} xdx = - [鈭� du 鈭� 2 鈭� u^{2} du + 鈭� u^{4} du] + C 鈭� \sin^{5} xdx = - [(\frac{u^{1}}{1}) 鈭� 2 (\frac{u^{2 + 1}}{2 + 1}) + (\frac{u^{4 + 1}}{4 + 1})] + C 鈭� \sin^{5} xdx = - [u 鈭� 2 (\frac{u^{3}}{3}) + (\frac{u^{5}}{5})] + C 鈭� \sin^{5} xdx = - [u 鈭� \frac{2}{3} u^{3} + \frac{1}{5} u^{5}] + C 鈭� \sin^{5} xdx = - u + \frac{2}{3} u^{3} - \frac{1}{5} u^{5} + C 鈭� \sin^{5} xdx = - \cos x + \frac{2}{3} \cos^{3} x - \frac{1}{5} \cos^{5} x + C$

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Short Answer

Step by step solution

Step 1. Given Information

Step 2. The given integral is ∫sin5xdx.

Step 3. Now the integral is

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