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

Evaluate the integral: $$\displaystyle\int{t}^{{{2}}}{{\sec}^{{{2}}}{\left({9}{t}^{{{3}}}+{1}\right)}}{\left.{d}{t}\right.}$$

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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, $$\displaystyle{\frac{{{1}}}{{{27}}}}{\tan{{\left({9}{t}^{{{3}}}+{1}\right)}}}+{C}$$.