Integrate: ∫ sin3 x cos2x dx

Dear student, Given that, ∫ sin3 x cos2x dx This can be written as: ∫ sin3 x cos2x dx = ∫ sin2 x cos2x (sin x) dx =∫(1 – cos2x ) cos2x (sin x) dx —(1) Now, substitute t = cos x, Then dt = -sin x dx Now, equation can be written as: Thus, ∫ sin3 x cos2x dx = – ∫ (1-t2)t2 dt Now, multiply t2 inside the bracket, we get = – ∫ (t2-t4) dt Now, integrate the above function: = – [(t3/3) – (t5/5)] + C —(2) Where C is a constant Now, substitute t = cos x in (2) = -(â…“)cos3x +(1/5)cos5x + C Hence, ∫ sin3 x cos2x dx = -(â…“)cos3x +(1/5)cos5x + C Regards

Integrate: ∫ sin³ x cos²x dx

Dear student,

Given that, ∫ sin³ x cos²x dx

This can be written as:

∫ sin³ x cos²x dx = ∫ sin² x cos²x (sin x) dx

=∫(1 – cos²x ) cos²x (sin x) dx —(1)

Now, substitute t = cos x,

Then dt = -sin x dx

Now, equation can be written as:

Thus, ∫ sin³ x cos²x dx = – ∫ (1-t²)t² dt

Now, multiply t² inside the bracket, we get

= – ∫ (t²-t⁴) dt

Now, integrate the above function:

= – [(t³/3) – (t⁵/5)] + C —(2)

Where C is a constant

Now, substitute t = cos x in (2)

= -(⅓)cos³x +(1/5)cos⁵x + C

Hence, ∫ sin³ x cos²x dx = -(⅓)cos³x +(1/5)cos⁵x + C
Regards

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