Problem on Definite Integration

Question

Q. Find all the values of $ 'a' $ which satisfy the equation: $$ ∫_{0}^{\frac{π}{2}} \left(\sin \left(x\right) + a\cos \left(x\right)\right)^3 dx - \frac{4a}{π-2} ∫_{0}^{\frac{π}{2}} x\cos \left(x\right) dx =2 $$

My Attempt: I tried to integrate the expression though I got the integral of $$ ∫_{0}^{\frac{π}{2}} x\cos \left(x\right) dx $$ But I am Clueless to integrate $$ ∫_{0}^{\frac{π}{2}} \left(\sin \left(x\right) + a\cos \left(x\right)\right)^3 dx $$ So please help me with that integral. Although I feel there must be another method where finding the integral is not necessary. Does any method exist? If it does, would you please share it to me ?

Any help would be appreciated

Answer 1

Note that by the change of variable $y=\dfrac\pi2-x$,

$$I(a):=\int_0^{\pi/2}(\sin x+a\cos x)^3dx=\int_0^{\pi/2}(\cos y+a\sin y)^3dy=a^3J(a^{-1}).$$

This shows that $I(a)$ is a cubic polynomial (by expanding the cube and integrating) with symmetric coefficients ($p+qa+qa^2+pa^3$) and you can spare two integral evaluations.

Among the terms of the development, $$p=\int_0^{\pi/2}\sin^3x\,dx=\int_0^{\pi/2}\sin x\sin^2x\,dx=\int_0^{\pi/2}\color{blue}{\sin x}\,dx-\int_0^{\pi/2}\color{blue}{\sin x\cos^2x}\,dx=1-\frac13=\frac23$$

and $$q=3\int_0^{\pi/2}\sin x\cos^2x\,dx=3\cdot\frac13$$ as already computed, which gives

$$I(a)=\frac{2+3a+3a^2+2a^3}3.$$

Now by parts,

$$\int_0^{\pi/2}x\color{blue}{\cos x}\,dx=\frac\pi2\sin\frac\pi2-\int_0^{\pi/2}\sin x\,dx=\frac\pi2-1.$$

The equation is

$$\frac{2+3a+3a^2+2a^3}3-2a=2.$$

By inspection, $$2a^3+3a^2-3a-4=0$$ has the solution $a=-1$, and by long division it factors as

$$(a+1)(2a^2+a-4).$$

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I don't think there is a way to avoid explicit computation of the coefficients, but a careful strategy reduces the integration effort to very little (actually just the three easy functions in blue).

Answer 2

Hint

If you expand $A=(\sin (x)+a \cos (x))^3$ you should get, $$A=\frac{1}{4} a^3 \cos ^3(x)+\frac{3}{4} a^3 \cos (x)-\frac{3}{4} a^3 \sin ^2(x) \cos (x)-\frac{3}{4} a^2 \sin ^3(x)+\frac{3}{4} a^2 \sin (x)+$$ $$\frac{9}{4} a^2 \sin (x) \cos ^2(x)-\frac{3}{4} a \cos ^3(x)+\frac{3}{4} a \cos (x)+\frac{9}{4} a \sin ^2(x) \cos (x)+$$ $$\frac{\sin ^3(x)}{4}+\frac{3 \sin (x)}{4}-\frac{3}{4} \sin (x) \cos ^2(x)$$ When integrating, the only problem is really with $\int \cos^3(x) \,dx$ and $\int \sin^3(x) \,dx$ but $$\int \cos^3(x) \,dx=\int \cos(x) (1-\sin^2(x)) \,dx$$ $$\int \sin^3(x) \,dx=\int \sin(x) (1-\cos^2(x)) \,dx$$ for which there is obvious subtitution.