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Parin Sir > Mathematical Proofs > Indefinite Integration > Pattern – 9 - Indefinite Integration > Pattern – 9 – Indefinite Integration

Pattern – 9 – Indefinite Integration

February 17, 2021
Category: Pattern – 9 - Indefinite Integration

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$\frac{1}{a\cos{x+b\sin{x}+c}}\$ $or\ \frac{1}{a\cos{x}+b\sin{x}}\$ $or\ \frac{1}{a\cos{x\ }+b}\$ $or\ \frac{1}{a\sin{x}+b}$

Let, $\tan{\frac{x}{2}}$ = t

∴ $\sec^2{\frac{x}{2}}$ × $\frac{1}{2}$ × dx = dt

$\therefore\ \ dx=\ \frac{2\ dt}{\sec^2{\frac{x}{2}}}$

$=\ \frac{2\ dt}{1+\tan^2{\frac{x}{2}}}$

$=\ \frac{2\ dt}{1+t^2}$

$\cos{x}=\ \frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}$

$=\ \frac{1-t^2}{1+t^2}$

$\sin{ x\ } = \ \frac{2\tan{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}$

$=\ \frac{2t}{1+t^2}$

$1) I=\ \int{\frac{1}{3+2\sin{x}+\cos{x}}\ dx}$

$Let,\tan{\frac{x}{2}}=t$

$\therefore \ dx = \frac{2dt}{1+t^2}$ &

3 + 2 sin x + cos x

$=3+\ 2\left(\frac{2t}{1+t^2}\right)+\ \left(\frac{1-t^2}{1+t^2}\right)$

$=\ \frac{3+3t^2+4t+1-t^2}{1+t^2}$

$=\ \frac{2t^2+4t+4}{1+t^2}$

$\therefore\ \ I=\ \int\frac{1+t^2}{2\ \left(t^2+2t+2\right)}\times\frac{2\ dt}{1+t^2}$

$=\ \int\frac{1}{t^2+2t+2\ }\ dt$

$=\ \int\frac{1}{\left(t^2+2t+1\right)+1}\ dt$

$=\ \int\frac{1}{\left(t+1\right)^2+1}\ dt$

$=\ \frac{1}{1}\tan^{-1}{\left(\frac{t+1}{1}\right)}+c$

$=\tan^{-1}{\left(1+\tan{\frac{x}{2}}\right)}+c$

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