∫tan53xsec63xdx=∫tan43xsec53xtan3xsec3xdx=∫{ ITEME1 sec53xtan3xsec3xdx=∫(ITEME2)sec53xtan3xsec3xdx=∫(ITEME3}sec53xtan3x
Posted: Thu Jul 14, 2022 4:03 pm
∫tan53xsec63xdx=∫tan43xsec53xtan3xsec3xdx=∫{ ITEME1 sec53xtan3xsec3xdx=∫(ITEME2)sec53xtan3xsec3xdx=∫(ITEME3}sec53xtan3xsec3xdx=∫sec93xtan3xsec3xdx−2∫sec73xtan3xsec3xdx+∫sec53xtan3xsec3xdx[ Let u=sec3x=∫u93du+(ITEME4)=10sec103x+(ITEME6}+{ITEME7)+C
∫cot5(32x)csc7(32x)dx=∫cot(32x)csc(32x)dx =∫(IEMF2)cot(32x)csc(32x)dx =∫(ITEMB3)csc6(32x)cot(32x)csc(32x)dx =∫[ [TIM F4] csc6(32x)cot(32x)csc(32x)dx =∫csc10(32x)cot(32x)csc(32x)dx+(ITEMF5)+∫csc6(32x)cot(32x)csc(32x)dx Let u=csc(32x) [ITM F6] =∫u10(2−3du)−2∫u8(2−3du)+∫u6(2−3du) =(2−3)(11u11)+2(23)(9u9)−(23)(7u7)+C ={TTEMF7}+{ITEMFB}+{ITEMF9}+C
∫sin33xcos53xdx=∫{ITEMD1}dx =∫({ITEMD2})cos53xsin3xdx =∫cos53xsin3xdx−∫cos73xsin3xdx Let u=cos3x dxdu={ITEMD3} {ITEMD4}=sin3xdx =∫u5(3−du)−∫u7(3−du) ={ITEMD5} ={ITEMD6}+{ITEMD7}+C
∫cot5(32x)csc7(32x)dx=∫cot(32x)csc(32x)dx =∫(IEMF2)cot(32x)csc(32x)dx =∫(ITEMB3)csc6(32x)cot(32x)csc(32x)dx =∫[ [TIM F4] csc6(32x)cot(32x)csc(32x)dx =∫csc10(32x)cot(32x)csc(32x)dx+(ITEMF5)+∫csc6(32x)cot(32x)csc(32x)dx Let u=csc(32x) [ITM F6] =∫u10(2−3du)−2∫u8(2−3du)+∫u6(2−3du) =(2−3)(11u11)+2(23)(9u9)−(23)(7u7)+C ={TTEMF7}+{ITEMFB}+{ITEMF9}+C
∫sin33xcos53xdx=∫{ITEMD1}dx =∫({ITEMD2})cos53xsin3xdx =∫cos53xsin3xdx−∫cos73xsin3xdx Let u=cos3x dxdu={ITEMD3} {ITEMD4}=sin3xdx =∫u5(3−du)−∫u7(3−du) ={ITEMD5} ={ITEMD6}+{ITEMD7}+C