# If I= int (sin 2x)/((3+4cosx)^(3))dx, then I equals

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Updated On: 3-5-2021

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Text Solution

(3cos x+8)/((3+4cosx)^(2))+C(3+8cos x)/(16(3+4cosx)^(2))+C(3+cos x)/((3+4cosx)^(2))+C(3-8cos x)/(16(3+4cosx)^(2))+C

B
Solution :
I= int (sin 2x)/((3+4cosx)^(3))dx <br> Put 3+4cosx=t," so that " -4sinx dx=dt. Then <br> I=(-1)/(8)int((t-3))/(t^(3))dt=(1)/(8)((1)/(t)-(3)/(2)(1)/(t^(2)))+C <br> =(2t-3)/(16t^(2))=(8cosx+3)/(16(3+4cosx)^(2))+C

Transcript

TimeTranscript
00:00 - 00:59welcome to doctor days is called the question I to integration sin 2x upon 3 + 4 cos x dx is equal to equal to integration to sin x cos x upon 1 + 4 cos x + 4 cos x = 25 - 4 sin x dx is equal to duty so we can purchase value above so it become to outside this is -1 by School this value becomes minus 3 by 4 upon this becomes
01:00 - 01:59- 12 integration t minus 3 upon 4 cube minus 1 by 2 integration of 1 by 4 p square DTE energy + 3 by 2 installation and bi fore u so this becomes minus 1 by 8 management + kids for minus 1 + 3 by 2 into 4 into 10 power minus 2 1 -2 School Tum so we have won by 80 upon 16 p square +
02:00 - 02:592 x minus 3 X square + so heavy put the value of cos x minus 16 cos square x + cos square x + 3 k we get the value of 4 cos x + 3 so we put it here it becomes 8cosx