What is the integral of 7^xcot^2(7^x) ?

The integral of 7^xcot^2(7^x) is 1/(ln(7))(-7^x-cot(7^x))+C

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integral of 7^xcot^2(7^x)

\int 7^{x} cot^{2} (7^{x}) d x

\frac{1}{ln ( 7 )} (- 7^{x} - cot (7^{x})) + C

Solution steps

\int 7^{x} cot^{2} (7^{x}) d x

Apply u-substitution

: \int \frac{cot ^{2} ( u )}{ln ( 7 )} d u

= \int \frac{cot ^{2} ( u )}{ln ( 7 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{ln ( 7 )} \cdot \int cot^{2} (u) d u

Rewrite using trig identities

= \frac{1}{ln ( 7 )} \cdot \int - 1 + csc^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{ln ( 7 )} (- \int 1 d u + \int csc^{2} (u) d u)

\int 1 d u = u

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{ln ( 7 )} (- u - cot (u))

Substitute back

u = 7^{x}

= \frac{1}{ln ( 7 )} (- 7^{x} - cot (7^{x}))

Add a constant to the solution

= \frac{1}{ln ( 7 )} (- 7^{x} - cot (7^{x})) + C

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What is the integral of 7^xcot^2(7^x) ?
The integral of 7^xcot^2(7^x) is 1/(ln(7))(-7^x-cot(7^x))+C