If x,dx^3x(1+^6x)^2/3=f(x)(1+^6x)^1/+c where c is a constant of integration, then f(3) is …

Question

If $\displaystyle\int\frac{\cos x\,dx}{\sin^3x(1+\sin^6x)^{2/3}}=f(x)(1+\sin^6x)^{1/\lambda}+c$ where c is a constant of integration, then $\lambda f\left(\dfrac{\pi}3\right)$ is equal to:

Accepted Answer

Let $\sin x=t\Rightarrow\cos x\,dx=dt$:
$$I=\int\frac{dt}{t^3(1+t^6)^{2/3}}=\int\frac{dt}{t^7\left(1+\frac1{t^6}\right)^{2/3}}$$

Put $1+\dfrac1{t^6}=r^3$; differentiating gives $-\dfrac6{t^7}dt=3r^2dr\Rightarrow \dfrac{dt}{t^7}=-\dfrac12r^2dr$:
$$I=-\frac12\int r^2\cdot r^{-2}dr=-\frac12 r+c=-\frac12\left(1+\frac1{t^6}\right)^{1/3}+c=-\frac{1}{2\sin^2x}(1+\sin^6x)^{1/3}+c$$

So $f(x)=-\dfrac12\csc^2x$ and $\lambda=3$. $\lambda f\left(\dfrac\pi3\right)=3\cdot\left(-\dfrac12\right)\csc^2\dfrac\pi3=3\cdot\left(-\dfrac12\right)\cdot\dfrac43=-2$

Solutions · 1