Evaluate the following integrals –
Let
Let us assume
We know and derivative of a constant is 0.
⇒ x + 3 = λ(0 – 4 – 2x2-1) + μ
⇒ x + 3 = λ(–4 – 2x) + μ
⇒ x + 3 = –2λx + μ – 4λ
Comparing the coefficient of x on both sides, we get
–2λ = 1
Comparing the constant on both sides, we get
μ – 4λ = 3
⇒ μ + 2 = 3
∴ μ = 1
Hence, we have
Substituting this value in I, we can write the integral as
Let
Now, put 3 – 4x – x2 = t
⇒ (–4 – 2x)dx = dt (Differentiating both sides)
Substituting this value in I1, we can write
Recall
Let
We can write 3 – 4x – x2 = –(x2 + 4x – 3)
⇒ 3 – 4x – x2 = –[x2 + 2(x)(2) + 22 – 22 – 3]
⇒ 3 – 4x – x2 = –[(x + 2)2 – 4 – 3]
⇒ 3 – 4x – x2 = –[(x + 2)2 – 7]
⇒ 3 – 4x – x2 = 7 – (x + 2)2
Hence, we can write I2 as
Recall
Substituting I1 and I2 in I, we get
Thus,