Question

Find the integrals of sin^2 (2x + 5)​

Answer 1

Answer:

Solution

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Consider the given integral.

I=∫sin

2

(2x+5)dx

Let t=2x+5

dx

dt

=2+0

2

dt

=dx

Therefore,

I=

2

1

∫sin

2

(t)dt

I=

4

1

∫(1−cos2t)dt

I=

4

1

[t−

2

sin2t

]+C

On putting the value of t, we get

I=

4

1

[2x+5−

2

sin2(2x+5)

]+C

I=

4

1

[2x+5−

2

sin(4x+10)

]+C

Hence, this is the answer.

Step-by-step explanation:

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