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huy266

Hai câu đầu giống nhau; làm câu 1 nhé:
I=∫cot⁡x.ln⁡3(sin⁡x)dx=∫ln⁡3(sin⁡x)sin⁡x.cos⁡xdxI=\int \cot x.\ln ^{3}(\sin x)dx=\int \frac{\ln ^{3}(\sin x)}{\sin x}.\cos xdx
Đặt u=sin⁡x⇒du=cos⁡xdxu=\sin x\Rightarrow du=\cos xdx
Vậy I=∫ln⁡3uudu=∫ln⁡3u.d(ln⁡u)=ln⁡4u4+CI=\int \frac{\ln ^{3}u}{u}du=\int \ln ^{3}u.d(\ln u)=\frac{\ln ^{4}u}{4}+C
I=ln⁡4(sin⁡x)4+CI=\frac{\ln ^{4}(\sin x)}{4}+C
 
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