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 December 31st, 2011, 10:24 PM #1 Newbie   Joined: Dec 2011 Posts: 25 Thanks: 0 integral of sin2x/2sincos^2x I'm studying integral calculus right now.. and my teacher gave as assignments.. I really don't have idea to answer this integral of sin2x/2sincos^2x but I try to answer this.. and I got I rewrite it to: 2sinxcosx/2sincos^2x then divide them and I got 1/cosx integral of sec x = ln |sec x + tan x| + c ln |sec x + tan x| + c this was my final answer.. is this correct? Happy new year and thank you in advance.
 December 31st, 2011, 10:38 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,193 Thanks: 504 Math Focus: Calculus/ODEs Re: integral of sin2x/2sincos^2x I assume you mean: $\int\frac{\sin2x}{2\sin x\cos^2x}\,dx=\int\frac{2\sin x\cos x}{2\sin x\cos^2x}\,dx=\int\sec x\,dx$ Let: $u=\sec x+\tan x\:\therefore\:du=\sec x$$\tan x+\sec x$$dx=u\sec x\,dx\:\therefore\:\sec x\,dx=\frac{1}{u}\,du$ and we have: $\int \frac{1}{u}\,du=\ln|u|+C=\ln\|\sec x+\tan x\|+C$ You are correct.
 December 31st, 2011, 10:53 PM #3 Newbie   Joined: Dec 2011 Posts: 25 Thanks: 0 Re: integral of sin2x/2sincos^2x Ohh my, am I really correct? Wow... Thank you very much.... as in super thank you She also gave this homework... integral of dx / 1-cosx I let u = sin x du =x dx=du integral of dx / u = dx/ sinx = 1/sinx = integral of csc x then my final answer was -ln|csc x + cot x | + c Am I also correct here? Thank you again...
December 31st, 2011, 10:58 PM   #4
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Re: integral of sin2x/2sincos^2x

Quote:
 Originally Posted by MarkFL I assume you mean: $\int\frac{\sin2x}{2\sin x\cos^2x}\,dx=\int\frac{2\sin x\cos x}{2\sin x\cos^2x}\,dx=\int\sec x\,dx$ Let: $u=\sec x+\tan x\:\therefore\:du=\sec x$$\tan x+\sec x$$dx=u\sec x\,dx\:\therefore\:\sec x\,dx=\frac{1}{u}\,du$ and we have: $\int \frac{1}{u}\,du=\ln|u|+C=\ln\|\sec x+\tan x\|+C$ You are correct.

Sir the denominator is only 2sincos^2x...

is it ok to divide with 2sinxcosx... and get 1 because there is x before cosx in numerator

 December 31st, 2011, 11:21 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,193 Thanks: 504 Math Focus: Calculus/ODEs Re: integral of sin2x/2sincos^2x We are given: $\int\frac{1}{1-\cos x}\,dx=\int\frac{1+\cos x}{\sin^2x}\,dx=\int\csc^2x+\cot x\csc x\,dx$ Observing that $-\frac{d}{dx}$$\cot x+\csc x$$=\csc^2x+\cot x\csc x$ we may write: $-\int\,d$$\cot x+\csc x$$=-$$\cot x+\csc x$$+C=-\cot$$\frac{x}{2}$$+C$
 January 1st, 2012, 12:05 AM #6 Newbie   Joined: Dec 2011 Posts: 25 Thanks: 0 Re: integral of sin2x/2sincos^2x You're so good... thank you again..
January 1st, 2012, 05:34 PM   #7
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Quote:
 Originally Posted by nephi39 . . . only 2sincos^2x...
That was presumably a typo and should have been 2sin x cos²x.

Using cos x = 1 - 2sin²(x/2),

$\int\frac{1}{1\,-\,\cos x}\,dx\,=\int\frac{1}{2\sin^2\frac{x}{2}}\,dx=\int \frac{\small1}{\small2}\csc^{\small2}\frac{x}{\sma ll2}\,dx\,=\,-\cot\frac{x}{\small2}\,+\;\text{C}.$

The substitution u = sin x (for which du = cos x dx) wouldn't have helped.

,

### 2sinco

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