Find the indefinite integral.a. ∫6x2(4x3-9)3dx b. ∫tan x sec2x dx

An Smith

An Smith

Answered question

2022-05-16

Find the indefinite integral.

a. 6x2(4x3-9)3dx

 

b. tan x sec2x dx

Answer & Explanation

alenahelenash

alenahelenash

Expert2022-06-05Added 556 answers

6x2(4x3-9)3dx

Since 6 is constant with respect to x, move 6 out of the integral.

6x2(4x3-9)3dx

Let u=4x3-9. Then du=12x2dx, so 112du=x2dx. Rewrite using u and du.

Let u=4x3-9. Find dudx.

12x2

Rewrite the problem using u and du.

61u3112du

Simplify.

Multiply 1u3 by 112.

61u312du

Move 12 to the left of u3.

6112u3du

Since 112 is constant with respect to u, move 112 out of the integral.

6(1121u3du)

Simplify the expression.

121u3du

Apply basic rules of exponents.

12u-3du

By the Power Rule, the integral of u-3 with respect to u is -12u-2.

12(-12u-2+C)

Simplify.

Rewrite 12(-12u-2+C) as 12(-121u2)+C.

12(-121u2)+C

Simplify.

-14u2+C

Replace all occurrences of uu with 4x3−94x3-9.

14(4x39)2+C

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