What is the integral of (sec^2(x))/(sqrt(25-tan^2(x))) ?

The integral of (sec^2(x))/(sqrt(25-tan^2(x))) is arcsin(1/5 tan(x))+C

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integral of (sec^2(x))/(sqrt(25-tan^2(x)))

\int \frac{sec ^{2} ( x )}{25 - tan ^{2} ( x )} d x

arcsin (\frac{1}{5} tan (x)) + C

Solution steps

\int \frac{sec ^{2} ( x )}{25 - tan ^{2} ( x )} d x

Apply u-substitution

: \int \frac{1}{25 - u ^{2}} d u

= \int \frac{1}{25 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

= 1 \cdot arcsin (\frac{1}{5} tan (x))

Simplify

= arcsin (\frac{1}{5} tan (x))

Add a constant to the solution

= arcsin (\frac{1}{5} tan (x)) + C

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What is the integral of (sec^2(x))/(sqrt(25-tan^2(x))) ?
The integral of (sec^2(x))/(sqrt(25-tan^2(x))) is arcsin(1/5 tan(x))+C