/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 39 Solve each of the integrals in E... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve each of the integrals in Exercises 39–74. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.

∫4-x2x2dx

Short Answer

Expert verified

The solution of the integral is-4-x2x-sin-1x2+C.

Step by step solution

01

Step 1. Given Information.

The given integral is∫4-x2x2dx.

02

Step 2. Solve.

To solve the integral, let x=2sinu,so the derivation of u isdx=2cosudu.

Thus, substitute u into the original integral.

localid="1648797152038" ∫4-x2x2dx=∫4-2sinu22sinu22cosudu=∫4-4sin2u4sin2u2cosudu=∫4(1-sin2u)4sin2u2cosudu=∫21-sin2u4sin2u2cosudu=∫1-sin2usin2ucosuduLet'susetheidentitysin2x+cos2x=1=∫cos2usin2ucosudu=∫cos2usin2udu

03

Step 3. Solve.

By proceeding with the calculation further,

=∫cos2usin2udu=∫cot2udu=∫csc2u-1du=∫csc2udu-∫1du=-cotu-u+C

Now, substitute back uin the above equation,

=-cotsin-1x2-sin-1x2+CUsetheidentitycotsin-1x=1-x2x=-1-x22x2-sin-1x2+C=-4-x2x-sin-1x2+C

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