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# Integration Techniques

Writing an integral down is only the first step. A toolkit of techniques can help find its value, from substitutions to trigonometry to partial fractions to differentiation.

\[\large \int_{\frac{1}{8}}^{\frac{1}{2}} \left \lfloor \ln \left \lceil \dfrac{1}{x} \right \rceil \right \rfloor dx = \, ?\]

\[\] **Notations**: \( \lfloor \cdot \rfloor \) denotes the floor function and \( \lceil \cdot \rceil \) denotes the ceiling function.

\[ \displaystyle \int_{1}^{\infty} \dfrac {2x\{x\} - \{x\}^2} {x^2 \lfloor x \rfloor ^2} \, dx \]

If the integral above can be expressed as \( \dfrac{\pi^a - b} c \), where \(a,b,c\) are all positive integers, find \(a+b+c\).

**Notations:**

- \( \{ \cdot \} \) denotes the fractional part function.
- \( \lfloor \cdot \rfloor \) denotes the floor function.

\[\large \int_0^\infty e^{-x}|\sin x| \, dx = \, ? \]

\[\large \displaystyle \int_0^\pi \dfrac{x\sin 2x \sin\left(\frac \pi 2 \cos x\right)}{2x-\pi} \, dx\]

The above integral can be expressed as \(\dfrac{A}{B\pi^C}\) where \(A\), \(B\) and \(C\) are positive integers with \(A\), \(B\) coprime. Find \(A+B+C\)

\[ \large \int_{0}^{\infty} \dfrac{\tan^{-1}(\pi x) - \tan^{-1}(x)}{x}\, dx = A\pi^{B}\dfrac{\ln \pi}{C} \]

The equation holds true for positive integers \(A,B,C\) with \(A,C\) coprime. Find \(A\times B\times C\).

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