Prove that: $\displaystyle\int_0^{\pi/4} \dfrac{\sin x+\cos x}{9+16\sin 2x}\,dx = \dfrac{1}{40}\log 9$ **OR** Prove that: $\displaystyle\int_0^{\pi} \dfrac{x\sin x}{1+\cos^2 x}\,dx = \dfrac{\pi^2}{4}$

> > [!TLDR] > Substitute $t=\sin x-\cos x$ so that $dt=(\sin x+\cos x)\,dx$ and $\sin2x=1-t^2$; the integral collapses to a standard $\int\frac{dt}{a^2-t^2}$ form that evaluates cleanly to $\frac{1}{40}\log9$. > > Let $t=\sin x-\cos x$. Then $dt=(\cos x+\sin x)\,dx$. > > Also, $t^2=\sin^2x-2\sin x\cos x+\cos^2x = 1-\sin2x$, so $\sin2x=1-t^2$. > > Denominator: $9+16\sin2x = 9+16(1-t^2)=25-16t^2$. > …

Prove that: ∫₀π/4 dfracsin x+cos x9+16sin 2xdx = d

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\displaystyle\int_0^{\pi/4} \dfrac{\sin x+\cos x}{9+16\sin 2x}\,dx = \dfrac{1}{40}\log 9

Tripura TbseHigher Secondary (+2 Stage) Examination 2023Subjective· 4mImportance★★★★★est

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