Integral Sin 3 X Dx. How do you evaluate the integral sin (x) sin (2x) sin (3x) dx? Let I=integ sinxsin2xsin3x dx or I=1/2×integ (cos2xcos4x)sin2x dx or I=1/4×integ2sin2xcos2x dx1/4× integ2cos4xsin2x dx or I=1/4× integ sin4x dx1/4×integ sin 6x dx+1/4×integ sin 2x dx orI=1/16cos4x+1/24 cos6x1/8cos2x Or1/48 (2cos 6x6cos 2x3cos 4x)+c 133 views.
Evaluateint (sin x)/(sin 3x) dxClass12Subject MATHSChapter INDEFINITE INTEGRAL BookARIHANT MATHS ENGLISHBoardIIT JEEYou can ask any doubt from class 61.
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Find the Integral sin(3x)dx Since is constant with respect to move out of the integral Integrate by parts using the formula where and Simplify Tap for more steps Combine and Combine and Combine and Since is constant with respect to move out of the integral Simplify.
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We can rewrite this as the indefinite integral Instead of sine squared x we’re saying sine is the same thing as u so we can rewrite that as u squared minus u to the fourth times du This is pretty straight forward now this is going to be u to the third over three minus u.
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How do you find the integral of sin 3 [x]dx? ∫sin 3 (x)dx Solution ∫sin 3 (x)dx=∫sin (x) (1−cos 2 (x))dx =∫sin (x)dx−∫sin (x)cos 2 (x)dx Now let us consider the first integral We know that ∫sin (x)dx=−cos (x)+C Now for the second integral We will use the substitution Let us assume that cos (x)=u Hence du=−sin (x)dx Therefore.