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Question

$1.\int\limits_{}^{} \frac{cosx}{\sqrt{cos2x}}dx$

$2.\int\limits_{}^{}\frac{2^x.3^x}{9^x-4^x}dx$

score 2 · Answer 1

$\int \dfrac{1}{(\dfrac{3}{2})^x - (\dfrac{2}{3})^x}dx =\int \dfrac{(\dfrac{3}{2})^x}{(\dfrac{3}{2})^{2x} -1}dx$

Đặt $(\dfrac{3}{2})^x = t \Rightarrow (\dfrac{3}{2})^x \ln \dfrac{3}{2} dx= dt$

$I= \dfrac{1}{\ln \dfrac{3}{2}} \int \dfrac{1}{t^2 -1}dt = \dfrac{1}{2\ln \dfrac{3}{2}} \int \bigg (\dfrac{1}{t-1}-\dfrac{1}{t+1} \bigg ) dt$

$= \dfrac{1}{2\ln \dfrac{3}{2}} \ln \bigg | \dfrac{t-1}{t+1} \bigg | + C =...$

score 1 · Answer 2

$\int \dfrac{\cos x}{\sqrt{1-2\sin^2 x}}dx =\int \dfrac{d(\sin x)}{\sqrt{1-2\sin^2 x}} =\int \dfrac{1}{\sqrt{1-2t^2}}dt$

Đặt $\sqrt 2 t = \sin u \Rightarrow \sqrt 2 dt = \cos u du$

$I=\dfrac{1}{\sqrt 2}\int \dfrac{\cos u}{\sqrt{1-\sin^2 u}}du =\dfrac{1}{\sqrt 2}\int du =\dfrac{1}{\sqrt 2} u + C$

$=\dfrac{1}{\sqrt 2} arc \sin \sqrt 2 t + C$

2 Đáp án