More practice questions

5. Q5. Evaluate ∫₀² dx/(x + 4 − x²) (NCERT Exercise 7.9, Q6, page 273).

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   x + 4 − x² = 17/4 − (x − ½)². Put t = x − ½, so the limits become −½ and 3/2, and use ∫dt/(a² − t²) with a = √17/2. The value is (1/√17) log ((21 + 5√17)/4).

6. Q6. Evaluate ∫₀^π/2 cos x/(1 + sin x) dx.

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   Put t = 1 + sin x; the limits become 1 and 2, and the integral is ∫₁² dt/t = log 2.

7. Q7. Evaluate ∫₁^e3 dx/(x√(1 + log x)).

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   Put t = 1 + log x; the limits become 1 and 4, and ∫₁⁴ t^−1/2 dt = 2(2 − 1) = 2.

8. Q8. Evaluate ∫₀^π/4 tan³ x sec² x dx.

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   Put t = tan x; the limits become 0 and 1, and the integral is ∫₀¹ t³ dt = ¼.

9. Q9. Evaluate ∫₀² x√(4 − x²) dx, and check that the answer is positive.

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   Put t = 4 − x², so x dx = −½ dt; x = 0 gives t = 4 and x = 2 gives t = 0. The integral is −½∫₄⁰ √t dt = ½∫₀⁴ √t dt = ½ × (2/3) × 8 = 8/3. It is positive, as it must be, because the integrand is not negative on [0, 2].

10. Q10. Evaluate ∫₀¹ e^{tan⁻¹ x}/(1 + x²) dx.

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    Put t = tan^⁻¹ x; the limits become 0 and π/4, and the integral is ∫₀^π/4 e^t dt = e^π/4 − 1.