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

The integral of sec^2(x)tan^7(x) is (tan^8(x))/8+C

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integral of sec^2(x)tan^7(x)

\int sec^{2} (x) tan^{7} (x) d x

\frac{tan ^{8} ( x )}{8} + C

Solution steps

\int sec^{2} (x) tan^{7} (x) d x

Rewrite using trig identities

= \int (1 + tan^{2} (x)) tan^{7} (x) d x

Apply u-substitution

: \int u^{7} d u

= \int u^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= \frac{u ^{8}}{8}

Substitute back

u = tan (x)

= \frac{tan ^{8} ( x )}{8}

Add a constant to the solution

= \frac{tan ^{8} ( x )}{8} + C

\int \frac{3 x ^{2} - 5 x + 1}{( x - 4 )} d x

e^{x^{5}}

f (x) = x^{3} - 3 x^{2}

\frac{d ^{2}}{d x ^{2}} (x + 3 x)

\int \frac{x ^{2} I n ^{3} ( 1 + x ^{3} )}{1 + x ^{3}} d x

What is the integral of sec^2(x)tan^7(x) ?
The integral of sec^2(x)tan^7(x) is (tan^8(x))/8+C