What is the integral of sqrt(1+cos(piθ)) ?

The integral of sqrt(1+cos(piθ)) is (2sqrt(2))/pi sin((piθ)/2)+C

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integral of sqrt(1+cos(piθ))

\int 1 + cos (π θ) d θ

\frac{2 2}{π} sin (\frac{π θ}{2}) + C

Solution steps

\int 1 + cos (π θ) d θ

Apply u-substitution

: \int 1 + cos (u) \frac{1}{π} d u

= \int 1 + cos (u) \frac{1}{π} d u

Take the constant out:

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

= \frac{1}{π} \cdot \int 1 + cos (u) d u

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{π} \cdot 22 \cdot \int \frac{1}{( 1 + v ^{2} ) 1 + v ^{2}} d v

Simplify

(1 + v^{2}) 1 + v^{2} : (1 + v^{2})^{\frac{3}{2}}

= \frac{1}{π} \cdot 22 \cdot \int \frac{1}{( 1 + v ^{2} ) ^{\frac{3}{2}}} d v

Apply Trigonometric Substitution

: \int cos (w) d w

= \frac{1}{π} \cdot 22 \cdot \int cos (w) d w

Use the common integral:

\int cos (w) d w = sin (w)

= \frac{1}{π} \cdot 22 sin (w)

Substitute back

= \frac{1}{π} \cdot 22 sin (arctan (tan (\frac{π θ}{2})))

Simplify

\frac{1}{π} \cdot 22 sin (arctan (tan (\frac{π θ}{2}))) : \frac{2 2}{π} sin (\frac{π θ}{2})

= \frac{2 2}{π} sin (\frac{π θ}{2})

Add a constant to the solution

= \frac{2 2}{π} sin (\frac{π θ}{2}) + C

\frac{x}{2}

\frac{2 s - 3}{2 s ^{2} - 8 s + 16}

f (x) = 4 x^{3} ln (x)

x \to 1 lim (tan (\frac{x}{2}))

y^{^{''}} - y = e^{t} (t + e^{t})

What is the integral of sqrt(1+cos(piθ)) ?
The integral of sqrt(1+cos(piθ)) is (2sqrt(2))/pi sin((piθ)/2)+C