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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.

Solveโˆซsin5xdx by using the Pythagorean identity sin2x+cos2x=1 to rewrite the integrand as role="math" localid="1649173494208" (1โˆ’cos2x)2sinx and then applying substitution with u=cosx.

Short Answer

Expert verified

The value of integral is โˆซsin5xdx=-cosx+23cos3x-15cos5x+C.

Step by step solution

01

Step 1. Given Information 

Solveโˆซsin5xdx by using the Pythagorean identity sin2x+cos2x=1 to rewrite the integrand as (1โˆ’cos2x)2sinx and then applying substitution with u=cosx.

02

Step 2. The given integral is โˆซsin5xdx.

We can write as

โˆซsin5xdx=โˆซ(1โˆ’cos2x)2sinxdx

Let

u=cosxdudx=-sinxdu=-sinxdx-du=sinxdx

03

Step 3. Now the integral is

โˆซsin5xdx=-โˆซ(1โˆ’u2)2duโˆซsin5xdx=-โˆซ{(1)2โˆ’2ร—1ร—u2+(u2)2}duโˆซsin5xdx=-โˆซ(1โˆ’2u2+u4)duโˆซsin5xdx=-โˆซduโˆ’2โˆซu2du+โˆซu4du+Cโˆซsin5xdx=-u11โˆ’2u2+12+1+u4+14+1+Cโˆซsin5xdx=-uโˆ’2u33+u55+Cโˆซsin5xdx=-uโˆ’23u3+15u5+Cโˆซsin5xdx=-u+23u3-15u5+Cโˆซsin5xdx=-cosx+23cos3x-15cos5x+C

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