![SOLVED: Using the identities cosh 2x + 1 cosh? x = cosh 2x - sinh? x = 2 sinh x cosh x = sinh 2x and the substitution u Zx, we can SOLVED: Using the identities cosh 2x + 1 cosh? x = cosh 2x - sinh? x = 2 sinh x cosh x = sinh 2x and the substitution u Zx, we can](https://cdn.numerade.com/ask_images/fa9b410087714a08b6d43ad9d662dc19.jpg)
SOLVED: Using the identities cosh 2x + 1 cosh? x = cosh 2x - sinh? x = 2 sinh x cosh x = sinh 2x and the substitution u Zx, we can
![STPM Further Mathematics T: 6.4 – Derivatives & Integrals of Hyperbolic & Inverse Hyperbolic Functions STPM Further Mathematics T: 6.4 – Derivatives & Integrals of Hyperbolic & Inverse Hyperbolic Functions](http://lh3.ggpht.com/-8gdAjF5T7tU/TgBq2rvoRNI/AAAAAAAAAV8/nUyLoi9lM_k/image_thumb5.png?imgmax=800)
STPM Further Mathematics T: 6.4 – Derivatives & Integrals of Hyperbolic & Inverse Hyperbolic Functions
For the integral [math] \int\sinh^3(x)dx [/math], why can't you just use substitution and let u = sinh(x)? - Quora
![integration - Integral of $e^{-k \cosh(z)} \text {sech}(z) \ dz$ from $z=0$ to $z=\infty$ - Mathematics Stack Exchange integration - Integral of $e^{-k \cosh(z)} \text {sech}(z) \ dz$ from $z=0$ to $z=\infty$ - Mathematics Stack Exchange](https://i.stack.imgur.com/9f0Zj.jpg)
integration - Integral of $e^{-k \cosh(z)} \text {sech}(z) \ dz$ from $z=0$ to $z=\infty$ - Mathematics Stack Exchange
![SOLVED: Section 6.9: Integrating Hyperbolic Functions Evaluate ∫ sinh(x) cosh(x) dx Evaluate ∫ dx / √x^2 + 16 Evaluate ∫ sin(θ) / √1 + cos^2(θ) dθ Evaluate ∫ √x dx Prove that SOLVED: Section 6.9: Integrating Hyperbolic Functions Evaluate ∫ sinh(x) cosh(x) dx Evaluate ∫ dx / √x^2 + 16 Evaluate ∫ sin(θ) / √1 + cos^2(θ) dθ Evaluate ∫ √x dx Prove that](https://cdn.numerade.com/ask_images/c765a361014e4954a4146eea5aa14965.jpg)