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integral of (cos(x))/((sin^2(x)+25)^2)

Solution

\int \frac{cos ( x )}{( sin ^{2} ( x ) + 25 ) ^{2}} d x

\frac{1}{250} (arctan (\frac{1}{5} sin (x)) + \frac{1}{2} sin (2 arctan (\frac{1}{5} sin (x)))) + C

Solution steps

\int \frac{cos ( x )}{( sin ^{2} ( x ) + 25 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{( u ^{2} + 25 ) ^{2}} d u

= \int \frac{1}{( u ^{2} + 25 ) ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{125 ( v ^{2} + 1 ) ^{2}} d v

Take the constant out:

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

= \frac{1}{125} \cdot \int \frac{1}{( v ^{2} + 1 ) ^{2}} d v

Apply Trigonometric Substitution

: \int \frac{1}{sec ^{4} ( w ) cos ^{2} ( w )} d w

= \frac{1}{125} \cdot \int \frac{1}{sec ^{4} ( w ) cos ^{2} ( w )} d w

Rewrite using trig identities

= \frac{1}{125} \cdot \int \frac{1 + cos ( 2 w )}{2} d w

Take the constant out:

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

= \frac{1}{125} \cdot \frac{1}{2} \cdot \int 1 + cos (2 w) d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{125} \cdot \frac{1}{2} (\int 1 d w + \int cos (2 w) d w)

\int 1 d w = w

\int cos (2 w) d w = \frac{1}{2} sin (2 w)

= \frac{1}{125} \cdot \frac{1}{2} (w + \frac{1}{2} sin (2 w))

Substitute back

= \frac{1}{125} \cdot \frac{1}{2} (arctan (\frac{sin ( x )}{5}) + \frac{1}{2} sin (2 arctan (\frac{sin ( x )}{5})))

Simplify

\frac{1}{125} \cdot \frac{1}{2} (arctan (\frac{sin ( x )}{5}) + \frac{1}{2} sin (2 arctan (\frac{sin ( x )}{5}))) : \frac{1}{250} (arctan (\frac{1}{5} sin (x)) + \frac{1}{2} sin (2 arctan (\frac{1}{5} sin (x))))

= \frac{1}{250} (arctan (\frac{1}{5} sin (x)) + \frac{1}{2} sin (2 arctan (\frac{1}{5} sin (x))))

Add a constant to the solution

= \frac{1}{250} (arctan (\frac{1}{5} sin (x)) + \frac{1}{2} sin (2 arctan (\frac{1}{5} sin (x)))) + C

\frac{\partial}{\partial y} (2 x y + y^{2})

x \frac{d y}{d x} + (- 1 + x) y = - x y^{2}

\int \frac{8}{x - 1} d x

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\frac{\partial}{\partial x} (\frac{5 x - y}{y ^{2} + 3 x ^{2}})