Q.Evaluate: $\int_0^{\pi/2} \dfrac{1}{6-\cos x}\,dx$

Most definite integrals in this chapter are evaluated by first finding an indefinite integral (antiderivative) using a suitable technique — algebraic substitution, trigonometric substitution (such as $x=a\sin\theta$ for $\sqrt{a^2-x^2}$-type integrands, or the Weierstrass substitution $t=\tan(x/2)$ for rational functions of $\sin x$ and $\cos x$), rationalising a surd denominator by multiplying by its conjugate, splitting a rational function by partial fractions, using standard trigonometric product-to-sum or power-reduction identities, or integration by parts — and then applying the Fundamental Theorem of Calculus (evaluate at the upper limit, subtract the value at the lower limit). The one extra care needed for a definite integral (as opposed to an indefinite one) is that when the integration variable is …