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Popular >

integral of 1/(sin(t)-cos(t))

Solution

\int \frac{1}{sin ( t ) - cos ( t )} d t

Solution

- 2 (\frac{1}{2} ln \frac{tan ( \frac{t}{2} ) + 1}{2} + 1 - \frac{1}{2} ln \frac{tan ( \frac{t}{2} ) + 1}{2} - 1) + C

Solution steps

\int \frac{1}{sin ( t ) - cos ( t )} d t

Apply u-substitution

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

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

Take the constant out:

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

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

Factor

2 u - 1 + u^{2} : - (- 2 u + 1 - u^{2})

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

Take the constant out:

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

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

Complete the square

- 2 u + 1 - u^{2} : - (u + 1)^{2} + 2

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

Apply u-substitution

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

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

Apply Integral Substitution

= 2 (- \int \frac{1}{2 ( - w ^{2} + 1 )} d w)

Take the constant out:

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

= 2 (- \frac{1}{2} \cdot \int \frac{1}{- w ^{2} + 1} d w)

Use the common integral:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= 2 (- \frac{1}{2} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= 2 - \frac{1}{2} \frac{ln \frac{t a n ( \frac{t}{2} ) + 1}{2} + 1}{2} - \frac{ln \frac{t a n ( \frac{t}{2} ) + 1}{2} - 1}{2}

Simplify

2 - \frac{1}{2} \frac{ln \frac{t a n ( \frac{t}{2} ) + 1}{2} + 1}{2} - \frac{ln \frac{t a n ( \frac{t}{2} ) + 1}{2} - 1}{2} : - 2 (\frac{1}{2} ln \frac{tan ( \frac{t}{2} ) + 1}{2} + 1 - \frac{1}{2} ln \frac{tan ( \frac{t}{2} ) + 1}{2} - 1)

= - 2 (\frac{1}{2} ln \frac{tan ( \frac{t}{2} ) + 1}{2} + 1 - \frac{1}{2} ln \frac{tan ( \frac{t}{2} ) + 1}{2} - 1)

Add a constant to the solution

= - 2 (\frac{1}{2} ln \frac{tan ( \frac{t}{2} ) + 1}{2} + 1 - \frac{1}{2} ln \frac{tan ( \frac{t}{2} ) + 1}{2} - 1) + C

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