Question

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Question in attachment u can solve any these questions and if possible solve all

Plz solve

Answer 1

you can see your answer

Answer 2

$\bold{Answer :}$

8.

Now,

∫ sin²x dx

= 1/2 ∫ (2 sin²x) dx

= 1/2 ∫ (1 - cos2x) dx [ ∵ 1 - 2 sin²x = cos2x ]

= 1/2 ∫ dx - 1/2 ∫ cos2x dx

= x/2 - 1/2 (1/2 sin2x) + c, where c is integral constant

= x/2 - (sin2x)/4 + c

9.

Now,

∫ cos²x dx

= 1/2 ∫ (2 cos²x) dx

= 1/2 ∫ (1 + cos2x) dx [ ∵ 2 cos²x - 1 = cos2x ]

= 1/2 ∫ dx + 1/2 ∫ sin2x dx

= x/2 + 1/2 ( - 1/2 cos2x) + c, where c is integral constant

= x/2 - (cos2x)/4 + c

10. (i)

Now,

∫ tan²x dx

= ∫ (sec²x - 1) dx [ ∵ sec²x - tan²x = 1 ]

= ∫ sec²x dx - ∫ dx

= tanx - x + c, where c is integral constant

(ii)

This one can be solved by taking 1/(cotx) = tanx and it will be same as 10. (i)

Now,

∫ (tanx/cotx) dx

= ∫ tan²x dx [ ∵ 1/(cotx) = tanx ]

= ∫ (sec²x - 1) dx [ ∵ sec²x - tan²x = 1 ]

= ∫ sec²x dx - ∫ dx

= tanx - x + c, where c is integral constant

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