I stumbled on this problem and I'm stuck.   Any hints on how I could approach evaluating this particular integral?  $$ \int_{-\pi/4}^{\pi/4}{ (\cos{t} + \sqrt{1 + t^2}\sin^3{t}\cos^3 {t})\,dt}$$  What I've tried prior to asking is using $\sin{t}\cos{t} = \frac{1}{2}\sin{2t}$ and $\sin^2{t} = 1 - \cos^2{t}$. I end up with:  $$ \int_{-\pi/4}^{\pi/4}{ (\cos{t} + \frac{1}{8}\sqrt{1 + t^2}\sin{2t} - \frac{1}{16}\sqrt{1 + t^2}\sin{4t}\cos{2t})\,dt}$$
40780
Evaluate: $$\int_{-\pi/4}^{\pi/4}(\cos(t)+\sqrt{1+t^2}\cos(t)^3\sin(t)^3)dt$$  I have tried substituting $t=\tan(\theta)$ and breaking up the $\cos(t)^3$ using $\cos(t)^2=1-\sin(t)^2$, but I can't figure it out.  This question is #21 from the Math subject GRE form GR1268.
13485
I'm stuck with this integral  $\int\frac {\cos^4x}{\sin^3x} dx$    which I rewrote as  $\int \csc^3x \cos^4xdx$  then after using the half angle formula twice for $\cos^4x$ I got this  $\frac 14\int \csc^3x (1+\cos(2x))(1+\cos(2x))dx$   then after solving those products I got these integrals  $\frac 14 \{\int \csc^3xdx+2\int \csc^3x \cos(2x)dx + \int \csc^3x \cos^2(2x)dx\}$  I do know how to solve the $\int \csc^3xdx$ one but I'm totally lost on the other ones, any tips/help/advice would be highly appreciated! thanks in advance guys!
79877
I have this difficult integral to solve.  $$ \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx} $$  Now my approach is this: split $(\sin x)^{\sqrt 2 + 1}$ and $(\sin x)^{\sqrt 2 - 1}$ as $(\sin x)^{\sqrt 2}.(\sin x)$ and $(\sin x)^{\sqrt 2 - 2}.(\sin x)$ respectively, and then apply parts. But that doesn't seem to lead anywhere. Hints please!  Edit:  This is what I did (showing just for the numerator)  $$ \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx $$ $$ = \int_0^{\pi/2} (\sin x)^{\sqrt 2}.(\sin x) dx  $$ $$ = (-\cos x)(\sin x)^{\sqrt 2}\Bigg|_0^{\pi/2} + \int_0^{\pi/2}(\sin x)^{ \sqrt 2 - 1 }(\cos^2 x) dx  $$  (taking $ v = \sin x $ and $ u = (\sin x)^{\sqrt 2} $ in the $ \int uv $ formula)  $$ = \int_0^{\pi/2}\left( (\sin x)^{ \sqrt 2 - 1 } - (\sin x)^{ \sqrt 2 + 1 } \right) dx  $$  Similarly for the denominator. This does give a reduction formula but then I don't see how to really use it for finding the answer.
85114
I'm having trouble computing the integral: $$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$ I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as $u=\sin(x)$ and $u=\cos(x)$, but it was not very effective.  Any suggestions are welcome. Thanks.
70985
How to evaluate the following integral $$\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$$ It looks like beta function but Wolfram Alpha cannot evaluate it. So, I computed the numerical value of integral above to 70 digits using Wolfram Alpha and I used the result to find its closed-form. The possible candidate closed-form from Wolfram Alpha is $$\pi\sqrt{\frac{1+\sqrt{2}}{2}}-\pi$$ Is this true? If so, how to prove it?
60496
How do we evaluate $\displaystyle\int_{0}^\pi \dfrac {\sin^2 x}{a^2-2ab \cos x + b^2}\mathrm dx$ ? I tried substitution and some other methods , but its not working ; please help .
72050
Evaluate the following integral:  $$\int \frac{\sqrt{\sin ^4x+\cos ^4x}}{\sin ^3x \cos x }dx$$ where $x \in \big(0,\frac{\pi}{2} \big)$  Could some give me hint as how to approach this question?  I tried to use the fact that $\sin ^4x+\cos ^4x=1-\frac{sin^22x}{2}$ but it didn't help. How should I proceed?
56896
Evaluate $$ \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin x} \text{d}x$$   where $n\in\mathbb{N}$     I was able to do this using parametization (differentiation under integration) but I'm not allowed to use that in my school since it isn't in the syllabus :( I should only use high school level mathematics. However, I can't see a way to do this with an elementary approach.
67059
The question :  $$ \int \frac {1-7\cos^2x} {\sin^7x \cos^2x} dx $$  I tried dividing by $\cos^2 x$ and splitting the fraction.  That turned out to be complicated(Atleast for me!)  How do I proceed now?
52543
I have no idea how to start, it looks like integration by parts won't work.  $$\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$$  If someone could shed some light on this I'd be very thankful.
118451
My question is, how can I evaluate the following integral?  $$\int\frac1{\sqrt[4]{1+x^4}}\mathrm dx$$  Thanks.
122468
My textbook says I should solve the following integral by first making a substitution, and then using integration by parts:  $$\int cos\sqrt x \ dx$$  The problem is, after staring at it for a while I'm still not sure what substitution I should make, and hence I'm stuck at the first step. I thought about doing something with the $\sqrt x$, but that doesn't seem to lead anywhere as far as I can tell. Same with the $cos$. Any hints?
41888
This problem is given on a sample test for my calculus two class. $$ \int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x}\,dx $$ I can find the value of this integral using other substitutions which lead to partial fractions but the prof added a hint to use the substitution $$t=\frac{\pi}{2}-x $$ so I've been trying to figure out how to do it his way but am pretty lost. Any ideas?
84518
Integrate the following  :   $$\int \frac{\sin x}{\sin4x}dx$$  My approach :   $$=\int \frac{\sin x}{2\sin2x \cos2x}dx$$  $$= \int \frac{\sin x}{4\sin x \cos x \cos2x}dx$$ $$= \int \frac{1}{4\cos x \cos2x}dx $$ [ didn't get the hint here ] then I did the following way.  $$= \int \frac{\sin x}{4\sin x \cos x (\cos2x) }dx$$  Now putting $\cos x =t$ $\Rightarrow -\sin x \, dx = dt $   $$= \int \frac{-dt}{4\sqrt{1-t^2} t(2t^2-1)}$$  [Using : $ \cos2x = 2\cos^2x-1]$
23617
So it's given this indefinite integral $$\int \frac{1}{\sin ^{\frac{1}{2}}x \cos^{\frac{7}{2}}x}dx$$  Is there anyone could solve this integral? Thanks in advance.
104093
$$\int{\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx}$$  $$\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx=\int \frac{(1-\sin x)(2-\sin x)}{\sqrt{(1-\sin x)(2-\sin x)(1+\sin x)(2+\sin x)}}dx$$  I am stuck. Please help me....
110923
$$\int^{ \pi /2}_{0} \ln (\sin x)\  dx$$  The answer is $- \frac{\pi}{2} \ln 2$.  I have changed it into $$\frac{1}{2} \int^{1}_{0} t d \ln t^{2}$$  But I didn't get the answer with it.