UPSC NDA Math Model Paper 10

I=∫

6

(sin‌x+sin‌2‌x)

dx
Using integration by parts
⇒I=∫

6

(sin‌x+2‌sin‌x‌cos‌x)

dx
⇒I=∫

6

sin‌x(1+2‌cos‌x)

dx
Multiply sin x both in numerator and denominator
⇒I=∫

6‌sin‌x

sin2x(1+2‌cos‌x)

dx
Substituting cos‌x=t⇒−sin‌x‌d‌x=dt
⇒I=∫

−6

(1−t)(1+t)(1+2t)

dt
By partial fraction
⇒I=−6‌∫

1

6(1−t)

−

1

2(1+t)

+

4

3(1+2t)

dt
⇒I=−6[

−1

6

‌ln|(1−t)|−

1

2

‌ln|(1+t)| +

4

3×2

‌ln|(1+2t)|]+C
⇒ I = ln | (1 − t) | + 3ln | (1 + t) | − 4ln | (1 + 2t) | + C
⇒ I = ln | 1 − cosx | + 3ln | 1 + cosx | − 4ln | 1 + 2cosx | + C