What is the integral of 10tan^9(x)sec^2(x) ?

The integral of 10tan^9(x)sec^2(x) is tan^{10}(x)+C

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integral of 10tan^9(x)sec^2(x)

\int 10 tan^{9} (x) sec^{2} (x) d x

tan^{10} (x) + C

Solution steps

\int 10 tan^{9} (x) sec^{2} (x) d x

Take the constant out:

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

= 10 \cdot \int tan^{9} (x) sec^{2} (x) d x

Rewrite using trig identities

= 10 \cdot \int (1 + tan^{2} (x)) tan^{9} (x) d x

Apply u-substitution

= 10 \cdot \int u^{9} d u

Apply the Power Rule

= 10 \cdot \frac{u ^{10}}{10}

Substitute back

u = tan (x)

= 10 \cdot \frac{tan ^{10} ( x )}{10}

Simplify

10 \cdot \frac{tan ^{10} ( x )}{10} : tan^{10} (x)

= tan^{10} (x)

Add a constant to the solution

= tan^{10} (x) + C

What is the integral of 10tan^9(x)sec^2(x) ?
The integral of 10tan^9(x)sec^2(x) is tan^{10}(x)+C