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\hspace{-16}%20$If%20$\vec{u}\times%20\vec{v}$%20be%20a%20Unit-Vector.%20If%20$\vec{w}$%20be%20a%20vector%20such%20that\\\\%20$\vec{w}\times%20u=\vec{v}-\vec{w}$%20and%20$\left|\left(\vec{u}\times%20\vec{v}\right).\vec{w}\right|=\frac{1}{2}$\\\\%20Then%20How%20can%20I%20prove%20$\vec{u}.\vec{v}=0$

\hspace{-16}\bf{\int%20\frac{1}{\ln(t+1)}dt}$\\\\\\%20Put%20$\bf{\ln(t+1)=x\Leftrightarrow%20t+1=e^x\Leftrightarrow%20dt=e^xdx}$\\\\\\%20So%20$\bf{\int%20\frac{e^x}{x}dx}$\\\\\\%20$\bf{\bullet%20\;e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+......\infty}$\\\\\\%20So%20$\bf{\int%20\frac{1+\frac{x}{1!}+\frac...

\hspace{-16}\bf{\int%20e^{4n}.(1-x)^3dx}$\\\\\\%20Using%20Integration%20By%20parts%20Recursively\\\\\\%20Where%20$\bf{(1-x)^2}$%20as%20$\bf{1^{st}}$%20and%20$\bf{e^{4n}}$%20as%20$\bf{2^{nd}}$\\\\\\%20So%20$\bf{\int%20e^{4n}.(1-x)^3dx=\frac{(1-x)^{3}.e^{4n}}{4}+\frac{3(1-x)^2.e^{4n}}{16}+\frac{6.(...

after editing

Thanks jyu but answer given is =1/5

\hspace{-16}\bf{\cos(3x)=4\cos^3(x)-3\cos%20(x)\Leftrightarrow%20\cos^3(x)=\frac{1}{4}\left\{\cos(3x)+3\cos(x)\right\}}$\\\\\\%20So%20$\bf{\left(\cos%20^3(x)\right)^2=\frac{1}{16}\left\{\cos^2(3x)+9\cos^2(x)+6\cos(3x).\cos%20(x)\right\}}$\\\\\\%20Now%20Using%20Half-Angle%20Formula\\\\\\%20$\bulle...

\hspace{-16}$Another%20easy%20Method%20for%20$\bf{\int%20\sin^5(x)dx}$\\\\\\%20$\bf{\int%20\sin^5(x)dx%20=\int%20\sin^4(x).\sin%20(x)dx}$\\\\\\%20$\bf{=\int%20\big(1-\cos%20^2(x)\big)^2.\sin%20(x)dx}$\\\\\\%20Let%20$\bf{\sin%20(x)=t\Leftrightarrow%20\cos%20(x)dx%20=dt}$\\\\\\%20$\bf{=\int%20(1-t^...

\hspace{-16}$%20Let%20$\bf{I_{n}=\int%20\sin^n(x)dx%20=\int\sin^{n-1}(x).\sin%20(x)dx}$\\\\\\%20Now%20Using%20Integration%20By%20Parts,%20Taking\\\\\\%20$\bf{\sin^{n-1}(x)}$%20as%20$\bf{1^{st}}$%20and%20$\bf{\sin%20(x)}$%20as%20$\bf{2^{nd}}$%20function.....\\\\\\%20So%20$\bf{I_{n}=-\sin^{n-1}(x)\...

\hspace{-16}$Here%20$\bf{\int\frac{1}{x^2.\sqrt{x-1}}dx}$\\\\\\%20Put%20$\bf{x=\frac{1}{t}\Leftrightarrow%20dx=-\frac{1}{t^2}dt}$\\\\\\%20so%20$\bf{-\int%20\frac{\sqrt{t}}{\sqrt{1-t}}dt}$\\\\\\%20Put%20$\bf{t=\sin^2%20\theta\Leftrightarrow%20dt=2\sin%20\theta.\cos%20\theta%20d\theta}$\\\\\\%20So%...

3 no. are choosen from {1,2,3,..........,8} with replacement, Then find the probability that min of choosen no. is 3 Given that max. of choosen no. is 6

Thanks Drongoski and Moderator actually it is from my assignment Thanks

Let f(x) be a function such that on putting any value of x we get the equation of a hyperbola which is conjugate of the hyperbola having x as eccentricity. then find, f(f(f(f(x)))

Thanks Mod. for Nice solution. But How can I prove the Given Result using Complex no. \bf{\frac{\sin^2(nx)}{\sin(x)} = \sum_{k=1}^{n} \sin \left [ (2k-1)x \right ]} Thanks

Thanks friends, yes answer is k=-1

thanks seanieg89

\bf{\int%20\frac{\sin^2(9x)}{\sin%20(x)}dx}

\bf{\int_{\pi /2}^{5\pi /2}{ \frac{e^{tan^{-1}(sinx)}}{e^{tan^{-1}(sinx) }+ e^{tan^{-1}(cosx)}}} dx} where \bf{\sin^{-1}(x) = arc (sin x)}

find value of k for which the equation x^4+(1-2k)x^2+(k^2-1) = 0 has one real solution

\hspace{-16}\bf{\int\frac{x^3-2}{\sqrt{(x^3+1)^3}}dx}$\\\\\\%20Using%20Hit%20and%20Trail......\\\\%20Let%20$\bf{\frac{d}{dx}\left\{\frac{Ax+B}{\sqrt{x^3+1}}\right\}=\frac{x^3-2}{\sqrt{(x^3+1)^3}}}$\\\\\\%20$\bf{\frac{2(x^3+1).A-(Ax+B).3x^2}{\sqrt{(x^3+1)^3}}=\frac{2x^3-4}{\sqrt{(x^3+1)^3}}}$\\\\\...

Thanks friends for answering. I soorry for my typo mistake,actually original question is \displaystyle \int \frac{x^3-2}{\sqrt{(x^3+1)^3}}dx