Few integrals

**mathwizard** · June 18th 2008, 10:55 AM

$\int \frac {sin(2x) - sin(2k)}{sin(x) - sin(k) + cos(x) - cos(k)}dx$

$<br /> \int \frac {dx}{\sqrt {cos^3(x + a)sin(x + b)}}$

$\int \sqrt {\frac {cos(x - a)}{cos(x + a)}} dx$

$\int \frac {cot(x) - 3cot(x)}{3tan(3x) - tan(x)} dx$

**galactus** · June 18th 2008, 11:29 AM

For the last one:

$\int\frac{-2cot(x)}{3tan(3x)-tan(x)}dx$

Rewrite as $\int\frac{4sin^{2}(x)-1}{4sin^{2}(x)}dx$

$=\int{dx}-\frac{1}{4}\int{csc^{2}(x)}dx$

$=x+\frac{1}{4}cot(x)+C$

**TheEmptySet** · June 18th 2008, 09:45 PM

Originally Posted by mathwizard

$\int \frac {sin(2x) - sin(2k)}{sin(x) - sin(k) + cos(x) - cos(k)}dx$

Using the sum to product identities we get

$\int \frac{2\cos(x+k)\sin(x-k)}{2\cos\left( \frac{x+k}{2}\right)\sin\left( \frac{x-k}{2}\right)-2\sin\left( \frac{x+k}{2}\right)\sin\left( \frac{x-k}{2}\right)}dx=$
$\int \frac{\cos(x+k)\sin(x-k)}{\sin\left( \frac{x-k}{2}\right)\left[\cos\left( \frac{x+k}{2}\right)-\sin\left( \frac{x+k}{2}\right)\right]}dx$

Now using the double angle identities on the numerator we get

$\int \frac{\left[\cos^2\left( \frac{x+k}{2}\right) -\sin^2\left( \frac{x+k}{2}\right) \right]2\sin\left(\frac{x-k}{2}\right)\cos\left( \frac{x-k}{2}\right)}{\sin\left( \frac{x-k}{2}\right)\left[\cos\left( \frac{x+k}{2}\right)-\sin\left( \frac{x+k}{2}\right)\right]}dx$

factoring and reducing we get

$\int 2\cos\left( \frac{x+k}{2}\right)\cos\left( \frac{x-k}{2}\right) -2\sin\left( \frac{x+k}{2}\right)\cos\left( \frac{x-k}{2}\right)dx$

Using the product to sum identities we get

$\int (\cos(x)+\cos(k)-(\sin(x)+\sin(k))dx=\sin(x)+x\cos(k)+\cos(x)-x\sin(k)+C$

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