Explained: "Evaluate the integral: ∫t2sec2(9t3+1)dt"

Reggie

Answered question

2021-11-07

Evaluate the integral:

\int t^{2} \sec^{2} (9 t^{3} + 1) d t

toroztatG

Skilled2021-11-08Added 98 answers

Step 1
The given integral is,

\int t^{2} \sec^{2} (9 t^{3} + 1) d t

.
Apply the u-substitution by taking

u = t^{3}

. Then

\frac{d u}{d t} = 3 t^{2}

. This implies that,

d t = \frac{1}{3 t^{2}} d u

\int t^{2} \sec^{2} (9 t^{3} + 1) d t = \int t^{2} \sec^{2} (9 u + 1) \frac{1}{3 t^{2}} d u

= \frac{1}{3} \int \sec^{2} (9 u + 1) d u

Step 2
Again, apply the u-substitution by taking v=9u+1. Then

\frac{d v}{d u} = 9

. This implies that,

d u = \frac{1}{9} d v

\int t^{2} \sec^{2} (9 t^{3} + 1) d t = \frac{1}{3} \int \sec^{2} (v) \frac{1}{9} d v

= \frac{1}{27} \int \sec^{2} (v) d v

= \frac{1}{27} \tan (v) + C

Substitute back, v=9u+1.

\int t^{2} \sec^{2} (9 t^{3} + 1) d t = \frac{1}{27} \tan (9 u + 1) + C

Substitute back,

u = t^{3}

\int t^{2} \sec^{2} (9 t^{3} + 1) d t = \frac{1}{27} \tan (9 t^{3} + 1) + C

Thus, the value of the integral is,

\frac{1}{27} \tan (9 t^{3} + 1) + C

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