Search results

After looking at my working I found a big brain substitution: \cos{x}=\frac{u}{\sqrt{1-u^2}} Which gets you straight to: \int \frac{1}{(1-2u^2)(u^2-1)} du so you can go partial fractions straight away.

Yep and you would have got what Vern and I got. The sub \cos{x}=\tan{u} does your two subs in one.

Yeah everything is good till the last step when you have to convert the u to a w so you will get a square root again.

Yeah honestly don't know what I was thinking, ironically the final answer is correct since the two negatives (which were both mistakes) cancel out.

Pretend you didnt see that ;). Also yes rookie mistake and fixed.

If anyone is bothered see if my result matches with vernburn's lol

Let \cos{x}=\tan{u} \implies-\sin{x} dx=\sec^2{u}du I=-\int \sqrt{\frac{1+\tan^2{u}}{(1-\tan^2{u})^2}} \times \sec^2{u}du \quad \text{since} \quad \sin{x}=\sqrt{1-\tan^2{u}} I=-\int \frac{\sec^3{u}}{1-\tan^2{u}} du I=-\int \frac{\frac{1}{\cos^3{u}}}{1-\frac{\sin^2u}{\cos^2{u}} du I=-...

Be back in 30mins once I type this up lol

? its cosec^2x and cot^2x not cosecx and cotx not to mention u need the dx

Yup nice now for a nasty one: \int{ \sqrt{\csc^2{x}+\cot^2{x}} dx

\int \frac{x^n}{ \sum_{k=0}^{n} \frac{x^k}{k!}}dx

This is a cool one: \int \left(\frac{x^2-3x+\frac{1}{3}}{x^3-x+1}\right)^2 dx

We can use the divide by x^2 on the top and bottom trick: \int \frac{x^2-1}{\left(x^2+1\right)\sqrt{x^4+1}}dx I=\int \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)\sqrt{\frac{x^4+1}{x^2}}}dx I=\int \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)\sqrt{(x+\frac{1}{x})^2-2}}dx Let...

Evaluate: \int_{0}^{\infty}\frac{\ln{x}}{1+x^2} dx

@CMtutor you know a smart way to solve this?

Yes well done guess it was too easy Try: \int \frac{1}{\sqrt[4]{x^4+1}} dx

Since this thread is dead I thought I would post a question: Find: \int \frac{1}{\sqrt{x\sqrt{x}-x^2}} dx

[is it meant to be sin^2{x} or sin^2{3x}? I dont see how you could solve the equation in it's current state (I tried product to sum, then triple angle identity but thats so grindy).

\frac{\cos{2x}-1}{x^3-x^2}=\frac{1-2\sin^2{x}-1}{x^3-x^2}=\frac{-2\sin^2{x}}{x^3-x^2}=\frac{-2\sin^2{x}}{x^3}+\frac{2\sin^2{x}}{x^2} The first limit approaches zero since its basically the limit of \frac{1}{x} and the second limit approaches 2 Edit: dumb sign mistake which is fixed.

I can almost guarantee you will get no students with that price...