tính

a, $\int\limits_{0}^{\frac{\pi}{2}}\cos^42xdx$

b, $\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{1+\sin 2x+\cos2x}{\sin x+\cos x}dx$

c, $\int\limits_{0}^{1}\frac{1}{e^x+1}dx$

d, $\int\limits_{0}^{\frac{\pi}{4}}\frac{\cos2x}{1+2\sin2x}dx$
$\int \dfrac{(1+\sin 2x) +\cos 2x}{\sin x + \cos x}dx =\int \dfrac{(\sin x +\cos x)^2 + (\cos x-\sin x)(\cos x + \sin x)}{\sin x +\cos x}dx$

$=\int (\sin x + \cos x)dx +\int (\cos x - \sin x)dx = 2\int \cos x dx = 2\sin x + C$

Câu a hạ bậc làm 1 lúc là ra, tự làm nha
$\int \dfrac{\cos 2x}{1+2\sin 2x}dx =\dfrac{1}{2}\int \dfrac{d(\sin 2x)}{1+2\sin 2x} =\dfrac{1}{2}\int \dfrac{1}{1+2t}dt$

$=\dfrac{1}{4} \int \dfrac{d(1+2t)}{1+2t} =\dfrac{1}{4} \ln |1+2t| + C$ tự đổi cận và thế vào nha
$I=\int \dfrac{e^x}{e^x(e^x+1)}dx =\int \dfrac{d(e^x)}{e^x(e^x+1)}=\int \dfrac{1}{t(t+1)}dt =\int \bigg (\dfrac{1}{t}-\dfrac{1}{t+1}\bigg )dt$

$=\ln|t| -\ln|t+1| + C = \ln \bigg |\dfrac{t}{t+1} \bigg | + C$

Bạn cần đăng nhập để có thể gửi đáp án

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