Evaluate:∫0π x9sin2⁡x+16cos2⁡xdx

Assume,I=∫0π x9sin2⁡x+16cos2⁡xdx…Now, recall that,∫ab f(x)dx=∫ab f(a+b−x)dxUsing that for given integral,∴I=∫0π π−x9sin2⁡(π−x)+16cos2⁡(π−x)dx∴I=∫0π π−x9sin2⁡x+16cos2⁡xdx…….(2)Adding equation (1) and (2)2I=∫0π π9sin2⁡x+16cos2⁡xdx⇒I=π∫0π2 19sin2⁡x+16cos2⁡xdx(using property)I=π∫0π/4 sec2⁡x9tan2⁡x+16dx+∫π/4π/2 cosec2⁡x9+16cot2⁡xdx⌋=πI1+I2 (say) =π∫01 dt9t2+16−∫10 dz9+16z2⊤ t=tan⁡x in⁡I1,z=cot⁡x in⁡I2=π12tan−1⁡3t4⌋01−tan−1⁡4z3011=π12tan−1⁡34+tan−1⁡43or π12×π2=π224Therefore, ∫0π x9sin2⁡x+16cos2⁡xdx=π12tan−1⁡34+tan−1⁡43. Which is the required answer.

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Evaluate:
$\int_{0}^{π} \frac{x}{9 \sin^{2} x + 16 \cos^{2} x} dx$

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Detailed Solution

Assume,
$I = \int_{0}^{π} \frac{x}{9 \sin^{2} x + 16 \cos^{2} x} dx \dots$
Now, recall that,
$\int_{a}^{b} f (x) dx = \int_{a}^{b} f (a + b - x) dx$
Using that for given integral,
$\begin{array}{l} ∴ I = \int_{0}^{π} \frac{π - x}{9 \sin^{2} (π - x) + 16 \cos^{2} (π - x)} dx \\ ∴ I = \int_{0}^{π} \frac{π - x}{9 \sin^{2} x + 16 \cos^{2} x} dx \dots \dots . (2) \end{array}$
Adding equation (1) and (2)
$\begin{array}{l} 2 I = \int_{0}^{π} \frac{π}{9 \sin^{2} x + 16 \cos^{2} x} dx \\ \Rightarrow I = π \int_{0}^{\frac{π}{2}} \frac{1}{9 \sin^{2} x + 16 \cos^{2} x} dx \end{array}$ (using property)
$\begin{array}{l} I = π [\int_{0}^{π / 4} \frac{\sec^{2} x}{9 \tan^{2} x + 16} dx + {\int_{π / 4}^{π / 2} \frac{{cosec}^{2} x}{9 + 16 \cot^{2} x} dx|}_{⌋} = π [I_{1} + I_{2}] (say) \\ = π [\int_{0}^{1} \frac{dt}{9 t^{2} + 16} - {\int_{1}^{0} \frac{dz}{9 + 16 z^{2}}|}^{⊤}⌋ (t = \tan x in I_{1}, z = \cot x in I_{2}) \\ = \frac{π}{12} \{\{{[{\tan^{- 1} \frac{3 t}{4}|}_{⌋_{0}}^{1} - {[\tan^{- 1} \frac{4 z}{3}]}^{0}_{1}]}_{1}) \\ = \frac{π}{12} (\tan^{- 1} \frac{3}{4} + \tan^{- 1} \frac{4}{3}) \end{array}$
or $\frac{π}{12} \times \frac{π}{2} = \frac{π^{2}}{24}$
Therefore, $\int_{0}^{π} \frac{x}{9 \sin^{2} x + 16 \cos^{2} x} dx = \frac{π}{12} (\tan^{- 1} \frac{3}{4} + \tan^{- 1} \frac{4}{3}) .$ Which is the required answer.