Question

intregration of sin^6xcos^7x from 0 to π/2​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

We have,

∫π20sin7xdx∫0π2sin7xdx

∫π20cos6x⋅sinxdx∫0π2cos6x⋅sinxdx

Time for assumption

Let us assume

cosx=t⟹−sinxdx=dtcos⁡x=t⟹−sin⁡xdx=dt

∫π20cosx6t6sinxdxdt∫0π2cosx6⏟t6sin⁡xdx⏟dt

change of limit

upper limit xxtends toπ4⟹ttends to12√tends toπ4⟹ttends to12

lower limitxtends to0⟹ttends to1xtends to0⟹ttends to1

-∫12√1t