How do you find int5/(16+9cos^2x)dx∫516+9cos2xdx? Calculus 2 Answers maganbhai P. Jul 22, 2018 I=1/4arc tan (4/5tanx)+cI=14arctan(45tanx)+c Explanation: Here , I=int5/(16+9cos^2x)dxI=∫516+9cos2xdx =int(5sec^2x)/(16sec^2x+9)dx.........to[becausecostheta=1/sectheta] =int(5sec^2x)/(16(1+tan^2x)+9)dxto[becausesec^2 theta-tan^2theta=1] =5intsec^2x/(16tan^2x+25)dx Subst. color(blue)(tanx=u=>sec^2xdx=du :.I=5int1/(16u^2+25)du =5/16int1/(u^2+25/16)du =5/16int1/(u^2+(5/4)^2)du =5/16*1/(5/4)arctan(u/(5/4))+c =5/16 xx 4/5arc tan((4u)/5)+c :.I=1/4arctan((4u)/5)+c Subst. back ,color(blue)(u=tanx I=1/4arc tan (4/5tanx)+c Answer link Ratnaker Mehta Jul 22, 2018 1/4arctan(4/5tanx)+C. Explanation: Let, I=int5/(16+9cos^2x)dx. :. I=5int1/{cos^2x(16/cos^2x+9)dx, =5int{(1/cos^2x)(1/(16sec^2x+9))}dx, =5int{1/{16(tan^2x+1)+9}}sec^2xdx. This suggests that the substn. 4tanx=u must work. Now, 4tanx=u rArr sec^2xdx=1/4du. :. I=5int{1/(u^2+5^2)}1/4du, =5/4*1/5arctan(u/5). rArr I=1/4arctan(4/5tanx)+C. Answer link Related questions How do I determine the molecular shape of a molecule? What is the lewis structure for co2? What is the lewis structure for hcn? How is vsepr used to classify molecules? What are the units used for the ideal gas law? How does Charle's law relate to breathing? What is the ideal gas law constant? How do you calculate the ideal gas law constant? How do you find density in the ideal gas law? Does ideal gas law apply to liquids? Impact of this question 2000 views around the world You can reuse this answer Creative Commons License