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

The integral of 10sec^2(x)tan(x) is 5tan^2(x)+C

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

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

5 tan^{2} (x) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int u d u

= 10 \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= 10 \cdot \frac{u ^{2}}{2}

Substitute back

u = tan (x)

= 10 \cdot \frac{tan ^{2} ( x )}{2}

Simplify

10 \cdot \frac{tan ^{2} ( x )}{2} : 5 tan^{2} (x)

= 5 tan^{2} (x)

Add a constant to the solution

= 5 tan^{2} (x) + C

\int x (1 + x^{2})^{11} d x

\int \frac{ln ^{2} ( x )}{x ^{5}} d x

\int (x^{5} - x^{3}) d x

\int x (x + 1)^{10} d x

\int (\frac{sin ( x )}{cos ( x )})^{2} d x

What is the integral of 10sec^2(x)tan(x) ?
The integral of 10sec^2(x)tan(x) is 5tan^2(x)+C