Evaluate the integral: \int t^{2}\sec^{2}(9t^{3}+1)dt

Reggie 2021-11-07 Answered
Evaluate the integral: \(\displaystyle\int{t}^{{{2}}}{{\sec}^{{{2}}}{\left({9}{t}^{{{3}}}+{1}\right)}}{\left.{d}{t}\right.}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

toroztatG
Answered 2021-11-08 Author has 14842 answers
Step 1
The given integral is, \(\displaystyle\int{t}^{{{2}}}{{\sec}^{{{2}}}{\left({9}{t}^{{{3}}}+{1}\right)}}{\left.{d}{t}\right.}\).
Apply the u-substitution by taking \(\displaystyle{u}={t}^{{{3}}}\). Then \(\displaystyle{\frac{{{d}{u}}}{{{\left.{d}{t}\right.}}}}={3}{t}^{{{2}}}\). This implies that, \(\displaystyle{\left.{d}{t}\right.}={\frac{{{1}}}{{{3}{t}^{{{2}}}}}}{d}{u}\).
\(\displaystyle\int{t}^{{{2}}}{{\sec}^{{{2}}}{\left({9}{t}^{{{3}}}+{1}\right)}}{\left.{d}{t}\right.}=\int{t}^{{{2}}}{{\sec}^{{{2}}}{\left({9}{u}+{1}\right)}}{\frac{{{1}}}{{{3}{t}^{{{2}}}}}}{d}{u}\)
\(\displaystyle={\frac{{{1}}}{{{3}}}}\int{{\sec}^{{{2}}}{\left({9}{u}+{1}\right)}}{d}{u}\)
Step 2
Again, apply the u-substitution by taking v=9u+1. Then \(\displaystyle{\frac{{{d}{v}}}{{{d}{u}}}}={9}\). This implies that, \(\displaystyle{d}{u}={\frac{{{1}}}{{{9}}}}{d}{v}\).
\(\displaystyle\int{t}^{{{2}}}{{\sec}^{{{2}}}{\left({9}{t}^{{{3}}}+{1}\right)}}{\left.{d}{t}\right.}={\frac{{{1}}}{{{3}}}}\int{{\sec}^{{{2}}}{\left({v}\right)}}{\frac{{{1}}}{{{9}}}}{d}{v}\)
\(\displaystyle={\frac{{{1}}}{{{27}}}}\int{{\sec}^{{{2}}}{\left({v}\right)}}{d}{v}\)
\(\displaystyle={\frac{{{1}}}{{{27}}}}{\tan{{\left({v}\right)}}}+{C}\)
Substitute back, v=9u+1.
\(\displaystyle\int{t}^{{{2}}}{{\sec}^{{{2}}}{\left({9}{t}^{{{3}}}+{1}\right)}}{\left.{d}{t}\right.}={\frac{{{1}}}{{{27}}}}{\tan{{\left({9}{u}+{1}\right)}}}+{C}\)
Substitute back, \(\displaystyle{u}={t}^{{{3}}}\).
\(\displaystyle\int{t}^{{{2}}}{{\sec}^{{{2}}}{\left({9}{t}^{{{3}}}+{1}\right)}}{\left.{d}{t}\right.}={\frac{{{1}}}{{{27}}}}{\tan{{\left({9}{t}^{{{3}}}+{1}\right)}}}+{C}\)
Thus, the value of the integral is, \(\displaystyle{\frac{{{1}}}{{{27}}}}{\tan{{\left({9}{t}^{{{3}}}+{1}\right)}}}+{C}\).
Not exactly what you’re looking for?
Ask My Question
0
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more
...