What is the integral of 1/(sec(x)sqrt(16-sin^2(x))) ?

The integral of 1/(sec(x)sqrt(16-sin^2(x))) is arcsin(1/4 sin(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sec(x)sqrt(16-sin^2(x)))

\int \frac{1}{sec ( x ) 16 - sin ^{2} ( x )} d x

arcsin (\frac{1}{4} sin (x)) + C

Solution steps

\int \frac{1}{sec ( x ) 16 - sin ^{2} ( x )} d x

Rewrite using trig identities

= \int \frac{cos ( x )}{16 - sin ^{2} ( x )} d x

Apply u-substitution

: \int \frac{1}{16 - u ^{2}} d u

= \int \frac{1}{16 - 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}{4} sin (x))

Simplify

= arcsin (\frac{1}{4} sin (x))

Add a constant to the solution

= arcsin (\frac{1}{4} sin (x)) + C

\int 4 sin (4 s) d s

\int \frac{5}{3 x} d x

\int \frac{8}{x ^{2} - 16} d x

\int (sin^{3} (x)) (cos^{4} (x)) d x

\int 7 (2 x - 9)^{6} d x

What is the integral of 1/(sec(x)sqrt(16-sin^2(x))) ?
The integral of 1/(sec(x)sqrt(16-sin^2(x))) is arcsin(1/4 sin(x))+C