What is the integral of 20tan^{19}(x)sec^2(x) ?

The integral of 20tan^{19}(x)sec^2(x) is tan^{20}(x)+C

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integral of 20tan^{19}(x)sec^2(x)

\int 20 tan^{19} (x) sec^{2} (x) d x

tan^{20} (x) + C

Solution steps

\int 20 tan^{19} (x) sec^{2} (x) d x

Take the constant out:

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

= 20 \cdot \int tan^{19} (x) sec^{2} (x) d x

Rewrite using trig identities

= 20 \cdot \int (1 + tan^{2} (x)) tan^{19} (x) d x

Apply u-substitution

: \int u^{19} d u

= 20 \cdot \int u^{19} d u

Apply the Power Rule

: \frac{u ^{20}}{20}

= 20 \cdot \frac{u ^{20}}{20}

Substitute back

u = tan (x)

= 20 \cdot \frac{tan ^{20} ( x )}{20}

Simplify

20 \cdot \frac{tan ^{20} ( x )}{20} : tan^{20} (x)

= tan^{20} (x)

Add a constant to the solution

= tan^{20} (x) + C

\int x - 1 - (x - 1)^{2} d x

\int \frac{x ^{3} + 3 x ^{2} - 18 x}{( x - 3 ) ( x + 6 )} d x

\int \frac{x + 2}{2} d x

\int \frac{20 ( 1 - x ^{2} )}{1 + x ^{2}} d x

What is the integral of 20tan^{19}(x)sec^2(x) ?
The integral of 20tan^{19}(x)sec^2(x) is tan^{20}(x)+C