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Exercise 4.2 · Q26

Q.Evaluate: ∫0π/4cos⁡x4−sin⁡2x dx\int_0^{\pi/4} \dfrac{\cos x}{4-\sin^2 x}\,dx

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t=sin⁡x, dt=cos⁡x dxt=\sin x,\ dt=\cos x\,dx; limits x=0→t=0x=0\to t=0, x=π/4→t=1/2x=\pi/4\to t=1/\sqrt2.

∫01/2dt4−t2=[14ln⁡2+t2−t]01/2.\int_0^{1/\sqrt2}\frac{dt}{4-t^2}=\Big[\frac14\ln\frac{2+t}{2-t}\Big]_0^{1/\sqrt2}. …

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