Solution:
∫xsecxtanx dx => ∣secx=cosx1∣ => ∫cos2xx⋅sinxdx (1)
After getting expression (1) we can use method integrate by parts: ∫v du=u⋅v−∫u dv
∫cos2xx⋅sinxdx=[v=x, dv=dx du=cos2xsinxdx,u∗=cosx1]=cosxx−∫cosxdx=cosxx−(∗∗)+C;
(∗∗) ∫cosxdx=∫cos2xcosxdx=∫cos2xd(sinx)=∫1−sin2xd(sinx) => ∣sinx=t ∣ => ∫1−t2dt
So, we get a tabular integrale: ∫1−t2dt=−21ln∣t+1t−1∣+C;
(∗∗) ∫cosxdx=−21ln∣sinx+1sinx−1∣;
∫cos2xx⋅sinxdx=cosxx+21ln∣sinx+1sinx−1∣+C;
u∗=∫cos2xsinxdx=−∫cos2xd(cosx)=cosx1;
Answer: ∫cos2xx⋅sinxdx=cosxx+21ln∣sinx+1sinx−1∣+C;
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